Moving Charges and Magnetic Field

a circle

Reason — When the electron enters perpendicularly, the angle between $\vec{\text v} \space$and $\vec{\text B} \space$is 90°, so the force F = Bqv sin 90° = Bqv is maximum and is always perpendicular to the velocity. Such a force does not change the speed of the particle, it only changes the direction of motion continuously. Hence it acts as the centripetal force and the electron describes a circular path of radius

$\text r = \frac{\text{mv}}{\text{qB}}$

magnetic flux density in CGS system

Reason — The SI unit of magnetic flux density $\vec{\text B} \space$is the tesla (T), that is, N A^-1 m^-1 or Wb m^-2. The gauss is the corresponding unit in the CGS system, and the two are related as

$1\ \text T = 10^4\ \text G$

kg s^-2 A^-1

Reason — From the defining relation for the magnetic field,

$\text B = \frac{\text F}{\text{qv} \sin \text θ}$

Taking F in newton (kg m s^-2), q in coulomb (A s) and v in m s^-1,

$$\begin{aligned}\text B = \frac{\text{kg m s}^{-2}}{(\text{A s})(\text{m s}^{-1})} \\[1em] \text B = \text{kg s}^{-2}\ \text A^{-1}\end{aligned}$$

velocity of the particle

Reason — For a charged particle injected normal to a uniform magnetic field, the period of revolution is

$\text T = \frac{2\text{πm}}{\text{qB}}$

This expression contains the mass, the charge and the magnetic field, but not the speed. If the speed increases, the radius also increases in the same proportion, so the time taken to complete one revolution remains the same. Hence the period is independent of the velocity of the particle.

Parabolic

Reason — A magnetic force always acts perpendicular to the instantaneous velocity; it cannot act as a constant force in a fixed direction, so it cannot produce a parabolic path.

Reason — For a long straight current-carrying conductor,

$$\begin{aligned}\text B = \frac{\text µ_0}{4\text π}\frac{2\text I}{\text a} \\[1em] \text B = \frac{\text µ_0}{2\text π}\frac{\text I}{\text a}\end{aligned}$$

so B is inversely proportional to the distance a of the point from the conductor. The graph of B against a is therefore a rectangular hyperbola, in which B falls off rapidly at first and then more and more slowly as a increases, never becoming zero. This variation is represented by the fourth graph.

downwards

Reason — Take east along $\hat{\text i}$, north along $\hat{\text j}$ and vertically upwards along $\hat{\text k}$. For the positive charge,

$\vec{\text F} = \text q(\vec{\text v} \times \vec{\text B})$

Here $\vec{\text v} = \text v\hat{\text i}$ and the deflecting force is along $\hat{\text j}$. If the field is directed vertically downwards, $\vec{\text B} = -\text B\hat{\text k}$, then

$$\begin{aligned}\vec{\text v} \times \vec{\text B} = \text{vB}\left(\hat{\text i} \times (-\hat{\text k})\right) \\[1em] \vec{\text v} \times \vec{\text B} = \text{vB}\ \hat{\text j}\end{aligned}$$

which is towards the north, as required. The same result follows from Fleming's left-hand rule.

90°

Reason — The magnitude of the magnetic force on a charged particle is

$\text F = \text{Bqv} \sin \text θ$

This is maximum when sin θ is maximum, that is when θ = 90°. Thus F_max = Bqv, obtained when the particle moves perpendicular to the field. At θ = 0 and θ = 180° the force is zero.

Li⁺⁺

Reason — Since all the particles are projected perpendicular to the field with the same velocity,

$\text F = \text{Bqv}$

so the force depends only on the magnitude of the charge. The charges are e for the electron, e for the proton, e for He⁺ and 2e for Li⁺⁺. Hence Li⁺⁺, carrying the largest charge, experiences the maximum force.

the momentum of the particle

Reason — For a charged particle moving perpendicular to a uniform magnetic field, the magnetic force provides the centripetal force,

$$\begin{aligned}\text{qvB} = \frac{\text{mv}^2}{\text r} \\[1em] \text r = \frac{\text{mv}}{\text{qB}} \\[1em] \text r = \frac{\text p}{\text{qB}}\end{aligned}$$

Thus the radius of curvature is directly proportional to the momentum p of the particle, and inversely proportional to the charge and to the intensity of the field.

both electric and magnetic fields

Reason — Every charge, whether at rest or in motion, produces an electric field around it. Oersted's experiment showed that a magnetic field is established around a current-carrying conductor, and since electric current is nothing but charge in motion, it is concluded that moving charges produce a magnetic field in the surrounding space. Hence a moving electric charge produces both an electric and a magnetic field.

both fields

Reason — Relative to the observer who moves past it, the electron is in motion. A charge in motion produces a magnetic field as well as an electric field. Hence the instrument carried by the moving observer measures both the fields. Only an observer at rest with respect to the electron would measure an electric field alone.

4B

Reason — Let the length of the wire be L. For one turn, L = 2πr₁, so $\text r_1 = \frac{\text L}{2\text π}$ and

$$\begin{aligned}\text B = \frac{\text µ_0 \text I}{2\text r_1} \\[1em] \text B = \frac{\text µ_0 \text I \text π}{\text L}\end{aligned}$$

For two turns, L = 2(2πr₂), so $\text r_2 = \frac{\text L}{4\text π}$ and

$$\begin{aligned}\text B' = \frac{\text µ_0 (2\text I)}{2\text r_2} \\[1em] \text B' = \frac{\text µ_0 \text I \times 4\text π}{\text L} \\[1em] \text B' = 4\text B\end{aligned}$$

$\sqrt{2}\text B$

Reason — The two coils have a common centre but their planes are perpendicular to each other. The magnetic field of a coil at its centre is perpendicular to the plane of the coil, so the two fields, each of magnitude B, are perpendicular to each other. The resultant is therefore

$$\begin{aligned}\text B_{\text{net}} = \sqrt{\text B^2 + \text B^2} \\[1em] \text B_{\text{net}} = \sqrt{2}\ \text B\end{aligned}$$

8 A

Reason — Given,

• B = 20 mT = 20 × 10^-3 T
• n = 20 turns per cm = 2000 turns per metre

For a long solenoid, B = µ₀nI, so

$$\begin{aligned}\text I = \frac{\text B}{\text µ_0 \text n} \\[1em] \text I = \frac{20 \times 10^{-3}}{(4\text π \times 10^{-7})(2000)} \\[1em] \text I = \frac{20 \times 10^{-3}}{2.51 \times 10^{-3}} \\[1em] \text I \approx 8\ \text A\end{aligned}$$

B ⊥ v

Reason — According to Biot-Savart law the field due to a current-element is

$\text{d}\vec{\text B} = \frac{\text µ_0}{4\text π}\frac{\text I(\vec{\text{dl}} \times \vec{\text r})}{\text r^3}$

so that $\text{d}\vec{\text B} \space$is perpendicular to the plane containing $\vec{\text{dl}} \space$and $\vec{\text r} \space$. As the direction of $\vec{\text{dl}} \space$is the direction of motion of the charges, $\vec{\text B} \space$is always perpendicular to $\vec{\text v} \space$. The law is also an inverse-square law, not an inverse-cube law.

1.5 × 10⁷ ms^-1

Reason — Given,

• E = 3 × 10⁴ volt/metre
• B = 2 × 10^-3 weber/metre²

The beam moves undeviated, so the electric force balances the magnetic force,

$$\begin{aligned}\text{qE} = \text{qvB} \\[1em] \text v = \frac{\text E}{\text B} \\[1em] \text v = \frac{3 \times 10^4}{2 \times 10^{-3}} \\[1em] \text v = 1.5 \times 10^7\ \text{ms}^{-1}\end{aligned}$$

$\sqrt{2}\text F$

Reason — The electron accelerated through a potential difference V acquires kinetic energy

$$\begin{aligned}\frac{1}{2}\text{mv}^2 = \text{eV} \\[1em] \text v = \sqrt{\frac{2\text{eV}}{\text m}}\end{aligned}$$

so that $\text v \propto \sqrt{\text V}$. Since the field is transverse, F = evB, giving $\text F \propto \sqrt{\text V}$. When V is increased to 2V,

$$\begin{aligned}\frac{\text F'}{\text F} = \sqrt{\frac{2\text V}{\text V}} \\[1em] \frac{\text F'}{\text F} = \sqrt{2} \\[1em] \text F' = \sqrt{2}\ \text F\end{aligned}$$

The electron will continue to move with uniform velocity along the axis of the solenoid

Reason — Inside a long solenoid the magnetic field is uniform and directed along the axis. The electron is projected along the same axis, so the angle between $\vec{\text v} \space$and $\vec{\text B} \space$is zero and

$$\begin{aligned}\text F = \text{Bqv} \sin 0° \\[1em] \text F = 0\end{aligned}$$

No force acts on the electron, so it continues to move undeviated with the same uniform velocity along the axis.

straight line

Reason — The particle starts from rest, so initially it experiences no magnetic force and is accelerated by the electric field alone, along the direction of $\vec{\text E} \space$. Since $\vec{\text B} \space$is parallel to $\vec{\text E} \space$, the velocity so acquired is also parallel to $\vec{\text B} \space$, and

$$\begin{aligned}\text F = \text{Bqv} \sin 0° \\[1em] \text F = 0\end{aligned}$$

The magnetic field therefore never exerts a force on it, and the particle continues to move in a straight line along the common direction of the two fields.

The charge to mass ratio satisfy : $\left(\frac{\text e}{\text m}\right)_1 + \left(\frac{\text e}{\text m}\right)_2 = 0$.

Reason — For a helical path the radius and the pitch are

$\text r = \frac{\text{mv} \sin \text θ}{\text{qB}}, \qquad \text P = \text v \cos \text θ \times \frac{2\text{πm}}{\text{qB}}$

The two paths are identical, so the two particles must have the same magnitude of the specific charge $\frac{\text q}{\text m}$. As the paths are traced in a completely opposite sense, the two charges must be of opposite sign. Both conditions together give

$\left(\frac{\text e}{\text m}\right)_1 + \left(\frac{\text e}{\text m}\right)_2 = 0$

This requires only equal specific charges of opposite sign, not equal charges or a particle-antiparticle pair.

undergoes acceleration all the time

Reason — Inside the dees the particle moves along a circular path, and although its speed is constant its direction changes continuously, so it possesses a centripetal acceleration. While crossing the gap between the dees it is accelerated by the electric field, and its speed increases. Hence the charged particle is under acceleration all the time. Inside a dee the particle is screened from the electric field by the metal walls of the dee, so it neither speeds up nor slows down there.

$\frac{25}{4}$

Reason — Let the length of the wire be L. For the coil of 2 turns, L = 2(2πr₁), so $\text r_1 = \frac{\text L}{4\text π}$ and

$$\begin{aligned}\text B_1 = \frac{\text µ_0 (2)\text I}{2\text r_1} \\[1em] \text B_1 = \frac{\text µ_0 \text I \times 4\text π}{\text L}\end{aligned}$$

For the coil of 5 turns, L = 5(2πr₂), so $\text r_2 = \frac{\text L}{10\text π}$ and

$$\begin{aligned}\text B_2 = \frac{\text µ_0 (5)\text I}{2\text r_2} \\[1em] \text B_2 = \frac{\text µ_0 \text I \times 25\text π}{\text L} \\[1em] \frac{\text B_2}{\text B_1} = \frac{25\text π}{4\text π} \\[1em] \frac{\text B_2}{\text B_1} = \frac{25}{4}\end{aligned}$$

the line integral of a magnetic field along the boundary of the closed loop is equal to µ₀ times the total current passing through the surface.

Reason — Ampere's circuital law states that the line integral of the magnetic flux density $\vec{\text B} \space$around any closed path in free space is equal to µ₀ times the net current I threading through the area enclosed by the path,

$\oint \vec{\text B} \cdot \vec{\text{dl}} = \text µ_0 \text I$

It is a line integral and not a surface integral, and only the current passing through the enclosed surface is counted, not a current passing near it.

B > A > C = D

Reason — By Ampere's circuital law,

$\oint \vec{\text B} \cdot \vec{\text{dl}} = \text µ_0 \text I$

so the line integral around a loop depends only on the net current threading that loop, and not at all on the shape or size of the loop or on the wires lying outside it. For a counterclockwise traversal, a current coming out of the page is taken as positive and one going into the page as negative.

Adding the enclosed currents with these signs for each loop of the figure, loop B encloses the largest net positive current, loop A the next, while loops C and D enclose equal net currents which are the smallest. Hence the ranking is B > A > C = D.

$\frac{\text µ_0 \text{Ia}}{2 \text{πb}^2}$

Reason — The current I is distributed uniformly over the cross-section of the wire of radius b. Applying Ampere's circuital law to a circle of radius a (a < b) drawn coaxially inside the wire, the current enclosed is

$$\begin{aligned}\text I_{\text{enc}} = \text I \times \frac{\text{πa}^2}{\text{πb}^2} \\[1em] \text I_{\text{enc}} = \frac{\text{Ia}^2}{\text b^2} \\[1em] \text B (2\text{πa}) = \text µ_0 \frac{\text{Ia}^2}{\text b^2} \\[1em] \text B = \frac{\text µ_0 \text{Ia}}{2\text{πb}^2}\end{aligned}$$

1420 Turns/m

Reason — Given,

• I = 2.0 A
• frequency of revolution, n = 1.0 × 10⁸ rev/s

The electron moves in a circle inside the solenoid, so its frequency of revolution is

$$\begin{aligned}\text n = \frac{\text{eB}}{2\text{πm}} \\[1em] \text B = \frac{2\text{πmn}}{\text e} \\[1em] \text B = \frac{2 \times 3.14 \times (9.1 \times 10^{-31}) \times (1.0 \times 10^8)}{1.6 \times 10^{-19}} \\[1em] \text B = 3.57 \times 10^{-3}\ \text T\end{aligned}$$

For the solenoid, B = µ₀NI where N is the number of turns per metre,

$$\begin{aligned}\text N = \frac{\text B}{\text µ_0 \text I} \\[1em] \text N = \frac{3.57 \times 10^{-3}}{(4\text π \times 10^{-7})(2.0)} \\[1em] \text N \approx 1420\ \text{turns/m}\end{aligned}$$

Negative z-axis

Reason — The electron carries a charge −e. The electric field is along the negative y-axis, so the electric force on the electron is

$$\begin{aligned}\vec{\text F}_{\text e} = (-\text e)(-\text E\hat{\text j}) \\[1em] \vec{\text F}_{\text e} = \text{eE}\ \hat{\text j}\end{aligned}$$

that is, along the positive y-axis. For the net force to be zero the magnetic force must be along the negative y-axis. With $\vec{\text v} = \text v\hat{\text i}$ and $\vec{\text B} = -\text B\hat{\text k}$,

$$\begin{aligned}\vec{\text F}_{\text m} = (-\text e)\left[\text v\hat{\text i} \times (-\text B\hat{\text k})\right] \\[1em] \vec{\text F}_{\text m} = (-\text e)(\text{vB}\ \hat{\text j}) \\[1em] \vec{\text F}_{\text m} = -\text{evB}\ \hat{\text j}\end{aligned}$$

which is along the negative y-axis, as required. Hence the magnetic field must be along the negative z-axis.

$\sqrt{\frac{2}{9}}\text R$

Reason — For a charged particle of kinetic energy K moving perpendicular to a magnetic field B,

$\text R = \frac{\sqrt{2\text{mK}}}{\text{qB}} \quad \Rightarrow \quad \text R \propto \frac{\sqrt{\text K}}{\text B}$

When the kinetic energy is doubled (K → 2K) and the magnetic induction tripled (B → 3B),

$$\begin{aligned}\frac{\text R'}{\text R} = \frac{\sqrt{2\text K}}{\sqrt{\text K}} \times \frac{\text B}{3\text B} \\[1em] \frac{\text R'}{\text R} = \frac{\sqrt{2}}{3} \\[1em] \frac{\text R'}{\text R} = \sqrt{\frac{2}{9}} \\[1em] \text R' = \sqrt{\frac{2}{9}}\ \text R\end{aligned}$$

helical

Reason — Here $\vec{\text B} = 10\hat{\text i}$ and $\vec{\text u} = 5\hat{\text i} + 4\hat{\text j}$. The velocity has a component $5\hat{\text i}$ parallel to the field and a component $4\hat{\text j}$ perpendicular to it. The parallel component is unaffected by the field and gives uniform motion along the field, while the perpendicular component gives a circular motion about the field direction. The superposition of the two is a helical path whose axis is parallel to $\vec{\text B} \space$.

$\frac{\text µ_0 4\text I}{\text{πL}}$

Reason —

The centre O of the square is at a perpendicular distance $\frac{\text L}{2}$ from each of the four infinitely long wires, so each wire produces at O a field of magnitude

$$\begin{aligned}\text B_1 = \frac{\text µ_0}{2\text π}\frac{\text I}{(\text L/2)} \\[1em] \text B_1 = \frac{\text µ_0 \text I}{\text{πL}}\end{aligned}$$

Applying the right-hand thumb rule to each wire in turn for the directions of current shown, all the four fields at O are directed perpendicular to the plane of the page, into the page. Hence they simply add up,

$$\begin{aligned}\text B = 4 \times \frac{\text µ_0 \text I}{\text{πL}} \\[1em] \text B = \frac{\text µ_0 4\text I}{\text{πL}}\end{aligned}$$

$\frac{\text µ_0 \text I}{2\text{πa}}$

Reason —

The point O lies on the line of the segments BO and OE, so for these segments the angle θ between the current-element and the line joining it to O is 0 or 180°, and they develop no magnetic field at O.

The remaining two segments are semi-infinite wires, each lying at a perpendicular distance a from O. For a semi-infinite straight conductor, φ₁ = 90° and φ₂ = 0, so each gives

$\text B_1 = \frac{\text µ_0}{4\text π}\frac{\text I}{\text a}$

By the right-hand thumb rule both these fields at O are in the same direction, so

$$\begin{aligned}\text B = 2 \times \frac{\text µ_0}{4\text π}\frac{\text I}{\text a} \\[1em] \text B = \frac{\text µ_0 \text I}{2\text{πa}}\end{aligned}$$

B₂ > B₁ for closer equidistant points only

Reason — Let the two wires be separated by a distance 2d and let P be a point on their perpendicular bisector at a distance R from each wire. Each wire alone gives

$\text B_1 = \frac{\text µ_0 \text I}{2\text{πR}}$

For anti-parallel currents the components of the two fields along the line joining the wires cancel, while the components perpendicular to it add. Resolving, the resultant is

$$\begin{aligned}\text B_2 = 2\text B_1 \cos \text θ \\[1em] \text B_2 = 2 \times \frac{\text µ_0 \text I}{2\text{πR}} \times \frac{\text d}{\text R} \\[1em] \text B_2 = \frac{\text µ_0 \text{Id}}{\text{πR}^2} \\[1em] \frac{\text B_2}{\text B_1} = \frac{2\text d}{\text R}\end{aligned}$$

Hence B₂ > B₁ only when R < 2d, that is, only at the equidistant points that are closer to the wires. For distant equidistant points R > 2d and the resultant becomes smaller than the field of a single wire.

The dimensions of $\frac{\text µ_0}{4\text π}$ are [M L T^-2 A^-2].

Explanation:
The unit of µ₀ is newton/ampere², so

$$\begin{aligned}\text µ_0 = \frac{\text{newton}}{\text{ampere}^2} \\[1em] \text µ_0 = \frac{[\text M \text L \text T^{-2}]}{[\text A^2]} \\[1em] \text µ_0 = [\text M \text L \text T^{-2} \text A^{-2}]\end{aligned}$$

Since 4π is a pure number, $\frac{\text µ_0}{4\text π}$ has the same dimensions.

The dimensions of $\frac{1}{\sqrt{\text µ_0 \text ε_0}}$ are [L T^-1].

Explanation:
The permeability and the permittivity of free space are related to the velocity of light in vacuum by

$$\begin{aligned}\text µ_0 \text ε_0 = \frac{1}{\text c^2} \\[1em] \frac{1}{\sqrt{\text µ_0 \text ε_0}} = \text c\end{aligned}$$

Hence this quantity has the dimensions of velocity, that is [L T^-1].

The magnitude of the magnetic field at the centre of a circular coil of N turns is

$\text B = \frac{\text µ_0 \text N \text I}{2 \text a}$

where a is the radius of the coil, I the current in it and µ₀ the permeability of free space.

Explanation:
The direction of $\vec{\text B} \space$is perpendicular to the plane of the coil, that is, along the axis of the coil.

Given,

• N = 50 turns
• a = 0.5 m
• I = 5 A

The magnetic flux density at the centre of a circular coil is

$\text B = \frac{\text µ_0 \text N \text I}{2 \text a}$

Substituting the values,

$$\begin{aligned}\text B = \frac{(4\text π \times 10^{-7})(50)(5)}{2 \times 0.5} \\[1em] \text B = \frac{4\text π \times 10^{-7} \times 250}{1} \\[1em] \text B = \text π \times 10^{-4}\ \text T\end{aligned}$$

Hence, the magnetic flux density at the centre of the coil is $\text π \times 10^{-4}$ T.

The particle will continue to move along a straight line, with no change in its speed or direction of motion.

The particle moves parallel to the field, so θ = 0 and

$$\begin{aligned}\text F = \text{Bqv} \sin 0° \\[1em] \text F = 0\end{aligned}$$

As no force acts upon it in the field, the particle moves undeviated.

The particle will move along a circle in a plane perpendicular to the field, with the same speed as before.

Here θ = 90°, so the particle experiences a maximum force F = Bqv, which is always perpendicular to its velocity. Such a force does no work on the particle, so the speed remains unchanged; it merely changes the direction of motion continuously and acts as the centripetal force.

The particle will move along a helical path whose axis is parallel to the magnetic field $\vec{\text B} \space$, with a constant speed.

Explanation:
The velocity is resolved into a component v cos θ parallel to $\vec{\text B} \space$and a component v sin θ perpendicular to it. The parallel component is unaffected by the field and gives a uniform linear motion along the field, while the perpendicular component gives a circular motion. The superposition of the two is a helical path.

The force on a current-element $\vec{\text{dl}} \space$carrying a current I placed in a magnetic field $\vec{\text B} \space$is

$\vec{\text F} = \text I\ \vec{\text{dl}} \times \vec{\text B}$

The direction of this force is perpendicular to the plane containing $\vec{\text{dl}} \space$and $\vec{\text B} \space$, that is, perpendicular to both of them. It may also be found by the right-hand palm rule no. 2 or by Fleming's left-hand rule.

No, a charge at rest experiences no force in a magnetic field, but it does experience a force in an electric field.

Explanation:
The magnetic force is F = Bqv sin θ, which is zero when v = 0. Only a charge in motion experiences a force in a magnetic field. The electric force, however, is F = qE, which is independent of the velocity, so a static charge does experience a force in an electric field.

(a) The pairs $(\vec{\text F}, \vec{\text v})$ and $(\vec{\text F}, \vec{\text B})$ are always at right angles.

(b) The pair $(\vec{\text v}, \vec{\text B})$ may have any angle between them.

Since $\vec{\text F} = \text q(\vec{\text v} \times \vec{\text B})$, the force is perpendicular to the plane containing $\vec{\text v} \space$and $\vec{\text B} \space$, and hence perpendicular to each of them. The angle θ between $\vec{\text v} \space$and $\vec{\text B} \space$itself may take any value, and it only decides the magnitude F = Bqv sin θ.

No, they do not represent the direction of the force on a moving charge.

Explanation:
The lines of force give the direction of $\vec{\text B} \space$at each point. The force on a moving charge is

$\vec{\text F} = \text q(\vec{\text v} \times \vec{\text B})$

which is perpendicular to $\vec{\text B} \space$, and hence perpendicular to the line of force at that point.

Here $\vec{\text v} \space$is along the +X-axis and $\vec{\text B} \space$along the +Y-axis, so θ = 90° and the magnitude of the force is

$\text F = \text{evB}$

For the direction, note that the electron is negatively charged, so the current is along the −X-axis, that is, opposite to the motion of the electron. Applying Fleming's left-hand rule with the current along −X and the field along +Y, the force on the electron is directed along the −Z-axis.

(i) The electron would be deflected towards the west.

(ii) If the electron is replaced by a proton, it would be deflected towards the east.

The electron moves vertically downwards and the field is directed from south to north. Since the electron is negatively charged, the direction of current is taken vertically upwards, and Fleming's left-hand rule then gives the force towards the west. For a proton the direction of current is the same as its direction of motion, so the force is reversed and is directed towards the east.

The beam will be deflected towards the north.

Explanation:
The electron-beam moves towards the east, so the direction of current is towards the west. With the field directed vertically upwards, Fleming's left-hand rule gives the force, and hence the deflection of the beam, towards the north.

The magnetic field in the room is directed vertically downwards.

Explanation:
The beam moves horizontally away from you, so the direction of current is towards you. The beam is deflected to your right. Applying Fleming's left-hand rule with the current directed towards you and the force towards your right, the fore-finger, which gives the direction of the field, points vertically downwards.

You are in the southern hemisphere.

Explanation:
The electron moves towards the north, so the direction of current is towards the south. The electron is deflected to your left, that is, towards the west. Applying Fleming's left-hand rule, the earth's magnetic field at that place must be directed vertically upwards. The earth's field points vertically upwards only in the southern hemisphere.

No, we cannot be sure that there is no magnetic field in that region.

Explanation:
A magnetic field may well be present, with the electron (or proton) moving parallel to it. In that case θ = 0, so

$$\begin{aligned}\text F = \text{Bqv} \sin 0° \\[1em] \text F = 0\end{aligned}$$

and the particle passes through undeflected.

No, it is not a definite indication of the presence of a magnetic field.

Explanation:
An electric field present in that region would also deflect the electron, since the electric force F = qE acts on a charge whether it is at rest or in motion. Hence the deflection alone cannot decide whether the field is magnetic or electric.

The conductor should be placed perpendicular to the field.

Explanation:
The force on a current-carrying conductor is

$\text F = \text{IB}l \sin \text θ$

which is maximum when sin θ = 1, that is when θ = 90°. Then F_max = IBl.

(a) When the currents are in the same direction, the magnetic field at the mid-point is zero.

(b) When the currents are in opposite directions, the field at the mid-point is twice the field due to either wire.

The mid-point is equidistant from the two wires, so the two fields are equal in magnitude. For currents in the same direction the two fields at the mid-point are opposite in direction and cancel, while for currents in opposite directions they are in the same direction and add up.

The magnetic field produced at the centre of the orbit is directed vertically downwards.

Explanation:
The electron is negatively charged, so the direction of the conventional current is opposite to its motion, that is, clockwise when seen from above. Applying the right-hand palm rule no. 1 to this clockwise current, the field at the centre of the orbit is directed vertically downwards.

Its direction of motion will look clockwise to us.

Explanation:
The field is directed vertically upwards and we are viewing from a height downwards, so the field points towards us. For a positive charge moving in a field directed towards the observer, the magnetic force $\vec{\text F} = \text q(\vec{\text v} \times \vec{\text B})$ turns the proton clockwise. (An electron in the same field would appear to move anticlockwise.)

Particle A is positively-charged, B is neutral and C is negatively-charged.

Explanation:
The field is perpendicular to the plane of the paper, pointing downwards (into the page). Particle B travels along a straight line, so no force acts on it and it must be uncharged. Particles A and C are deflected in opposite senses, so they carry charges of opposite sign; applying Fleming's left-hand rule to the direction of bending shown, A is positively charged and C is negatively charged.

The initial velocity $\vec{\text v} \space$must be either parallel or anti-parallel to $\vec{\text B} \space$.

Explanation:
The particle travels undeflected with a constant speed, so the field exerts no force on it. Since

$$\begin{aligned}\text F = \text{Bqv} \sin \text θ \\[1em] \text F = 0\end{aligned}$$

and neither B nor v is zero, sin θ must be zero, that is θ = 0 or 180°. The variation of the magnitude of the field from point to point does not matter, because the direction of $\vec{\text B} \space$is constant.

The radius of the proton's path is $\sqrt{1835}$ times larger than that of the electron.

For a particle accelerated through a potential difference V and entering the field perpendicularly,

$$\begin{aligned}\text r = \frac{\sqrt{2\text{mqV}}}{\text{qB}} \\[1em] \text r = \frac{1}{\text B}\sqrt{\frac{2\text{mV}}{\text q}}\end{aligned}$$

The charges are equal in magnitude and V and B are the same, so $\text r \propto \sqrt{\text m}$. Hence

$$\begin{aligned}\frac{\text r_\text p}{\text r_\text e} = \sqrt{\frac{\text m_\text p}{\text m_\text e}} \\[1em] \frac{\text r_\text p}{\text r_\text e} = \sqrt{1835}\end{aligned}$$

The ratio of the radii of their paths is 4 : 1.

Both are α-particles, so they have the same mass and the same charge, and they enter the same field perpendicularly. From

$\text r = \frac{\text{mv}}{\text{qB}}$

r is directly proportional to v. Hence

$$\begin{aligned}\frac{\text r_\text A}{\text r_\text B} = \frac{\text v_\text A}{\text v_\text B} \\[1em] \frac{\text r_\text A}{\text r_\text B} = \frac{4}{1}\end{aligned}$$

Biot-Savart's law : The magnetic field dB developed at a point P due to a small current-element I dl of a current-carrying conductor is

(i) directly proportional to the current I flowing through the conductor,

(ii) directly proportional to the length dl of the element,

(iii) directly proportional to the sine of the angle θ between the current-element and the line joining the element to the point P, and

(iv) inversely proportional to the square of the distance r between the element and the point P.

Combining these,

$\text{dB} = \frac{\text µ_0}{4\text π}\frac{\text I\ \text{dl} \sin \text θ}{\text r^2}$

In vector form,

$\text d\vec{\text B} = \frac{\text µ_0}{4\text π}\frac{\text I(\vec{\text{dl}} \times \vec{\text r})}{\text r^3}$

where µ₀ is the permeability of free space.

The magnetic lines of force around a straight, long current-carrying conductor are concentric circles lying in a plane perpendicular to the conductor, with their centres on the conductor. The lines are crowded close to the wire and become more widely spaced farther from it, showing that the field weakens with distance.

Right-hand thumb rule : If we grasp the straight conductor in the right hand such that the thumb points in the direction of the current, then the curled fingers give the direction of the magnetic field $\vec{\text B} \space$, with the finger tips pointing along $\vec{\text B} \space$.

(i) When the magnet is cut perpendicular to its length : Each piece is a complete magnet with a north and a south pole, having the same pole-strength as before but a reduced length.

(ii) When the magnet is cut along its length : Each piece is again a complete magnet with a north and a south pole, having the same length as before but a reduced pole-strength.

In either case an isolated pole is never obtained, because magnetic monopoles do not exist.

No, it is not necessary for every magnetic field configuration to have a north pole and a south pole.

The poles exist only when the source of the magnetic field has a net magnetic moment. A toroid has zero magnetic moment, so it does not have north and south poles. Its magnetic lines of force lie entirely within the core in the form of concentric circles, closing on themselves, and the field outside the core is zero.

The force on a charge q moving with velocity $\vec{\text v} \space$in a uniform magnetic field $\vec{\text B} \space$is

$\vec{\text F} = \text q(\vec{\text v} \times \vec{\text B})$

Its magnitude is

$\text F = \text{Bqv} \sin \text θ$

where θ is the angle between $\vec{\text v} \space$and $\vec{\text B} \space$.

The force is zero when θ = 0 or 180°, that is, when the charge moves parallel or anti-parallel to the field. It is also zero when the charge is at rest (v = 0).

If the given field is a uniform electric field, the electric force qE on the electrons is constant in magnitude and direction, so the path of the beam will be parabolic.

If it is a uniform magnetic field, the force is always perpendicular to the velocity, so the path will be circular when the beam enters perpendicular to the field, and helical when it enters at some other angle.

Hence the nature of the path traced by the deflected beam tells us which of the two fields is present.

Solenoid : A solenoid is a long, closely wound helical coil of insulated wire, whose length is very large compared with its diameter. When a current is passed through it, it behaves like a bar magnet.

The lines of force inside the solenoid are straight, parallel and equidistant, showing a uniform magnetic field, except near the edges. Outside the solenoid the lines spread out and the field is very small compared with the field inside. One end of the solenoid behaves as a north pole and the other as a south pole.

For a point well inside a long current-carrying solenoid,

$\text B = \text µ_0 \text n \text I$

and at a point at one end of the solenoid,

$\text B_{\text{end}} = \frac{1}{2}\text µ_0 \text n \text I$

where n is the number of turns per metre length of the solenoid and I is the current in it. Thus the field at the end is one half of the field at the middle.

Given,

• q = 10 µC
• $\vec{\text v} = (4\hat{\text i} + 5\hat{\text j})$ m/s
• $\vec{\text F} = (5\hat{\text i} - \text C\hat{\text j})$ N

The force on a moving charge in a magnetic field is

$\vec{\text F} = \text q(\vec{\text v} \times \vec{\text B})$

which is always perpendicular to $\vec{\text v} \space$. Hence

$$\begin{aligned}\vec{\text F} \cdot \vec{\text v} = 0 \\[1em] (5\hat{\text i} - \text C\hat{\text j}) \cdot (4\hat{\text i} + 5\hat{\text j}) = 0 \\[1em] 20 - 5\text C = 0\end{aligned}$$

Hence, C = 4.

Fleming's left-hand rule : Stretch the fore-finger, the middle-finger and the thumb of the left hand so that these are perpendicular to one another. If the fore-finger points in the direction of the magnetic field and the middle-finger in the direction of the current, then the thumb indicates the direction of the magnetic force.

For a positively charged particle the direction of motion is taken as the direction of current, while for a negatively charged particle the direction of current is taken opposite to its direction of motion.

(a) When the currents are in the same direction, the wires attract each other.

(b) When the currents are in opposite directions, the wires repel each other.

Each wire lies in the magnetic field produced by the current in the other, and so experiences a force given by F = I₂B₁L. The force per unit length between the two wires is

$\frac{\text F}{\text L} = \frac{\text µ_0}{2\text π}\frac{\text I_1 \text I_2}{\text r}$

and its direction, found by the right-hand palm rule, is towards the other wire for parallel currents and away from it for anti-parallel currents.

Each current-carrying wire lies in the magnetic field produced by the current in the other wire.

Consider the wire PQ carrying a current I₁. The magnetic field due to it at the position of the second wire RS is perpendicular to the plane of the paper. The wire RS, carrying the current I₂ in the opposite direction, is thus placed in this field and experiences a magnetic force

$\text F = \text I_2 \text B_1 \text L \sin 90°$

Applying Fleming's left-hand rule, this force is directed away from PQ. By the same argument the force on PQ is directed away from RS. Hence the two wires repel each other.

Definition of ampere : One ampere is that steady current which, when maintained in each of two infinitely long, straight and parallel conductors of negligible cross-section placed in vacuum at a distance of 1 metre from each other, produces a force of 2 × 10^-7 newton per metre length on each conductor.

This follows from the force per unit length between two parallel conductors carrying the same current I,

$\frac{\text F}{\text L} = \frac{\text µ_0}{2\text π}\frac{\text I^2}{\text r}$

Putting I = 1 A and r = 1 m gives $\frac{\text F}{\text L} = 2 \times 10^{-7}$ N m^-1. Since the currents are in the same direction, this force is attractive.

The needle will point along the north-east direction, instead of north.

The electron-beam moves from south to north, so the direction of the current is from north to south. By the right-hand thumb rule, the magnetic field of this current at a point just below the beam is directed towards the east.

The needle is therefore acted upon by two fields at right angles, the earth's horizontal field towards the north and the field of the beam towards the east. It sets itself along the resultant of the two, that is, towards the north-east.

The wire buried in the wall is vertical, and the current in it flows from bottom to top.

Since the north pole of the needle points towards the south, the magnetic field of the wire at the position of the needle is directed towards the south. As the needle lies in a horizontal plane, the field of the wire must lie in the horizontal plane, and this is possible only if the wire is vertical.

It is clear from the figure that the lines of force due to the current-carrying wire are anti-clockwise. So, by the right-hand thumb rule, the current in the wire should be from bottom to top.

The magnetic field at the centre of the cube is zero.

The twelve wires of the cube form six pairs of currents, namely AB and HG, AE and CG, AD and FG, BF and DH, FE and CD, and EH and BC. The two currents of each pair are equally inclined and equally distant from the centre but are oppositely directed, so they produce equal and opposite magnetic fields at the centre of the cube. Hence the fields of every pair cancel out and the resultant field at the centre is zero.

(a) No magnetic field is produced at the centre O.

All the four wires are of the same material and of the same cross-sectional area, so they have equal resistances and equal currents flow through them. The fields at O due to the currents in AB and DC (and similarly due to those in BC and AD) are equal in magnitude but opposite in direction, so they cancel mutually. Hence there is no magnetic field at O.

(b) A magnetic field is produced at the centre O.

Aluminium has a higher resistivity than copper, so the aluminium wires AD and DC carry smaller currents than the copper wires BC and AB. The cancellation is therefore incomplete, and a magnetic field exists at O due to the excess currents in BC and AB. This field is perpendicular to the plane of the page, directed downwards (right-hand palm rule no. 1).

(i) Yes, there is a magnetic field inside. For a point inside the wire at a distance R from the axis,

$\text B = \frac{\text µ_0 \text I}{2\text π}\frac{\text R}{\text a^2}$

so the field is directly proportional to the distance R of the point from the centre of the wire.

(ii) Yes, there is a magnetic field outside. For a point outside,

$\text B = \frac{\text µ_0 \text I}{2\text{πR}}$

so the field is inversely proportional to R.

Note: The question first speaks of a solid copper wire and then of a pipe. The answer above is for a solid wire, as taken in the textbook. For a hollow pipe the field at an inside point would be zero, since no current would then be enclosed.

The magnetic field at the centre of the coil is

$\text B = \frac{\text µ_0 \text N \text I}{2 \text a}$

(i) If the current is reduced to half, then since B ∝ I, the field will be halved.

(ii) If the radius is doubled, then since $\text B \propto \frac{1}{\text a}$, the field will again be halved.

(iii) The current is clockwise when looked at from above, so by the right-hand palm rule no. 1 the field at the centre is perpendicular to the plane of the paper, pointing downwards.

The field will be stronger in the second case, and four times as strong as in the first case.

Let l be the length of the wire. The field at the centre of a coil of N turns is

$\text B = \frac{\text µ_0 \text N \text I}{2 \text a}$

For one turn, l = 2πa₁, so $\text a_1 = \frac{l}{2\text π}$ and

$\text B_1 = \frac{\text µ_0 \text I}{2(l/2\text π)}$

For two turns, l = 2(2πa₂), so $\text a_2 = \frac{l}{4\text π}$ and

$$\begin{aligned}\text B_2 = \frac{\text µ_0 (2\text I)}{2(l/4\text π)} \\[1em] \frac{\text B_2}{\text B_1} = 4\end{aligned}$$

No force is produced on either conductor.

The wire carrying the current I₂ lies along the axis of the circular conductor, and the axial magnetic lines of force due to the current I₁ are along this very axis. As the wire coincides with the direction of the field, θ = 0 and it experiences no force due to I₁.

Similarly, the circular conductor carrying the current I₁ coincides with the circular lines of force produced by I₂, so it too experiences no force due to I₂.

In both the cases the lines of force are circular and anti-clockwise.

By Maxwell's right-hand screw rule, taking the direction of motion of a positive charge as the direction of current, a positive charge coming directly towards you is equivalent to a current directed towards you, and the lines of force around it are anti-clockwise.

A negative charge going directly away from you is equivalent to a current directed towards you as well, so here also the lines of force are circular and anti-clockwise.

(i) Both the electrons will describe similar circular paths, and both will move clockwise [Fig. (a)]. They have the same charge and the same speed in a field of the same magnitude, so their radii are equal, but since they start in opposite directions their circles lie on opposite sides of the starting point.

(ii) The electron and the positron carry charges of opposite sign, so the magnetic force on them is oppositely directed. They will move on the same circle, the electron clockwise and the positron anti-clockwise. They will therefore meet again and annihilate each other [Fig. (b)].

Because the particles lose energy continuously in the liquid, so the radius of the track goes on decreasing.

A charged particle ionises the medium much more in a liquid than in a gas. The particles therefore progressively lose energy and are slowed down by collisions with the environment.

The radius of the track in a magnetic field is

$\text r = \frac{\text{mv}}{\text{qB}}$

which is proportional to the speed of the particle. As the speed decreases progressively, the radius also goes on decreasing, and the circular track becomes a spiral track.

The speed, and hence the kinetic energy, of the particle does not change; the momentum does change; and in an electric field the kinetic energy does change.

The magnetic force on a charged particle moving in a uniform magnetic field is always perpendicular to its direction of motion, so no work is done by the field on the particle and its speed (or kinetic energy) remains unchanged. The direction of motion, however, changes continuously, so the momentum of the particle changes.

In an electric field the force qE on the charged particle is along the direction of the field and not perpendicular to the velocity. Work is therefore done on the particle and its kinetic energy changes.

The particle leaves the field with its initial speed, but in a different direction.

The magnetic force is always perpendicular to the velocity of the particle, whatever be the magnitude and direction of $\vec{\text B} \space$at each point. Hence the field does no work on the particle and the magnitude of the velocity (that is, the speed) remains unchanged throughout; only the direction of the velocity goes on changing.

The magnetic force is related to the motion of charge, and not to the net charge of the conductor.

In a current-carrying conductor the free electrons move with the drift velocity v_d, so each of them experiences a magnetic force in the field. The positive ions of the lattice are stationary and so they do not experience any magnetic force, although their charge cancels that of the electrons and makes the net charge of the conductor zero.

The resultant of these microscopic forces on the drifting electrons is the macroscopic force

$\text F = \text{IB}l \sin \text θ$

experienced by the whole conductor.

Like the molecules of a gas, the free electrons in a conductor are in a state of continuous random motion. They therefore have no net velocity in any direction, and the magnetic forces on the individual electrons, being in all possible directions, cancel out. Hence there is no net magnetic force on the conductor.

When a current is passed, the electrons acquire a drift velocity v_d in a definite direction in addition to their random motion, and the magnetic forces on them now add up to give a resultant force on the conductor.

The loop would move towards the wire AB.

Let the rectangular loop be named PQRS, with the arm SP nearer to AB. The currents in AB and SP are parallel, so there is an attraction between them, while the currents in AB and QR are anti-parallel, so there is a repulsion between them.

The force per unit length between two parallel conductors is

$\frac{\text F}{\text L} = \frac{\text µ_0}{2\text π}\frac{\text I_1 \text I_2}{\text r}$

which is inversely proportional to the separation r. As SP is nearer to AB than QR, the attraction is larger than the repulsion. Hence there is a net attraction and the loop moves towards AB.

The body will be lifted upwards, and this does not depend upon the direction of the current in the spring.

The current in every turn of the spring is in the same direction, so any two neighbouring turns behave like two parallel conductors carrying currents in the same direction, and therefore attract each other. The turns consequently come closer, the spring contracts, and the suspended body is lifted upwards.

On reversing the current the currents in all the turns are reversed together, so they are still parallel and the attraction remains. Hence the effect is the same whatever be the direction of the current.

The spring will execute vibrations.

With an alternating current the magnitude of the current in the turns changes periodically, and so does the force of attraction between any two turns, which is proportional to the product of the currents. The force therefore varies periodically between zero and a maximum value.

The spring accordingly contracts and expands periodically, that is, it executes vibrations.

The electrons of the beam repel one another electrically, which makes the beam diverge. But the electrons are also moving charges in the same direction, and so they behave like parallel currents, which attract one another magnetically.

At ordinary velocities the electrical repulsion is much stronger than the magnetic attraction, so the beam diverges. As the velocity is raised, the magnetic attraction becomes more effective. At very high velocities the beam starts converging.

For the movable wire XY to remain stationary, the magnetic force F on it due to the current i₂ in the fixed wire PQ must balance its weight Mg,

$\text F = \text{Mg}$

In the original setup PQ lies below XY. To balance the downward weight, the magnetic force on XY must be upward, away from PQ; therefore it is a force of repulsion, and the two currents i₁ and i₂ are in opposite directions.

When XY is placed below the fixed wire PQ, the force on XY must again be directed upwards, but this is now towards PQ, so it must be an attraction. Since parallel currents attract, the current i₁ in XY must now be in the same direction as i₂ in PQ, and the redrawn figure is labelled accordingly.

Biot-Savart's law : Let XY be a conductor of arbitrary shape carrying a current I, and P a point at which the field is required. Let I dl be a current-element of the conductor and r the distance of this element from P, θ being the angle between the element and the line joining it to P.

According to Biot-Savart law, the magnetic field dB at the point P due to the current-element I dl is

(i) proportional to the current, dB ∝ I,

(ii) proportional to the length of the element, dB ∝ dl,

(iii) proportional to sin θ, and

(iv) inversely proportional to the square of the distance, dB ∝ 1/r².

Combining these four rules,

$\text{dB} = \frac{\text µ_0}{4\text π}\frac{\text I\ \text{dl} \sin \text θ}{\text r^2}$

and in vector form

$\text d\vec{\text B} = \frac{\text µ_0}{4\text π}\frac{\text I(\vec{\text{dl}} \times \vec{\text r})}{\text r^3}$

Here I is the current in the conductor, dl the length of the current-element, r the distance of the point from the element, θ the angle between the element and the line joining it to the point, and $\frac{\text µ_0}{4\text π}$ the proportionality constant, µ₀ being the permeability of free space, equal to 4π × 10^-7 N A^-2.

The field due to the entire conductor is obtained by integrating over its whole length,

$$\begin{aligned}\vec{\text B} = \int \text d\vec{\text B} \\[1em] \vec{\text B} = \frac{\text µ_0 \text I}{4\text π}\int \frac{\vec{\text{dl}} \times \vec{\text r}}{\text r^3}\end{aligned}$$

Definition of tesla : The unit of magnetic field obtained from this law is newton/(ampere-metre), which is called the tesla (T). One tesla is that magnetic flux density which exerts a force of 1 newton on a conductor of length 1 metre carrying a current of 1 ampere, placed perpendicular to the field.

Relation between gauss and tesla : The gauss is the corresponding unit in the CGS system, and

$1\ \text{tesla} = 10^4\ \text{gauss}$

Analogies (similarities) :

(i) Coulomb's law gives the electric field due to a distribution of charges, while Biot-Savart's law gives the magnetic field due to a current-element. Biot-Savart's law is thus the magnetic equivalent of Coulomb's law.

(ii) According to Coulomb's law the magnitude of the electric field at a point P due to a charge-element dq is

$\text{dE} = \frac{1}{4\text{πε}_0}\frac{\text{dq}}{\text r^2}$

and according to Biot-Savart law the magnitude of the magnetic field at P due to a current-element I dl is

$\text{dB} = \frac{\text µ_0}{4\text π}\frac{\text I\ \text{dl} \sin \text θ}{\text r^2}$

Both have a constant of proportionality depending on the medium.

(iii) Both are inverse-square laws, the field falling off as $\frac{1}{\text r^2}$ in each case.

Differences (dissimilarities) :

(i) The charge-element dq is a scalar, whereas the current-element $\text I \vec{\text{dl}} \space$is a vector whose direction is along the direction of the current.

(ii) According to Coulomb's law the magnitude of the electric field depends only upon the distance of the charge-element from the point P. According to Biot-Savart's law the magnitude of the magnetic field at P depends also upon the angle θ between the current-element and the line joining the element to the point P. The field is zero along the direction of the current (θ = 0 or 180°) and maximum perpendicular to it (θ = 90°).

(iii) According to Coulomb's law the direction of the electric field is along the line joining the charge-element and the point P. According to Biot-Savart's law the direction of the magnetic field is perpendicular to the current-element as well as to the line joining the current-element to the point P.

Ampere's circuital law : The line integral of the magnetic flux density $\vec{\text B} \space$around any closed path in free space is equal to µ₀ times the net current I threading through the area enclosed by the path,

$\oint \vec{\text B} \cdot \vec{\text{dl}} = \text µ_0 \text I$

Magnetic flux density near a long current-carrying conductor :

Let a long straight conductor carry a steady current I, and let X be a point at a perpendicular distance r from it, at which the field is required.

By symmetry, the magnitude of $\vec{\text B} \space$is the same at all points of a circle of radius r drawn around the wire in a plane perpendicular to it, and $\vec{\text B} \space$is tangential to this circle at every point. Choosing this circle as the closed path, $\vec{\text B} \space$and $\vec{\text{dl}} \space$are along the same direction at each element, so

$$\begin{aligned}\oint \vec{\text B} \cdot \vec{\text{dl}} = \oint \text B\ \text{dl} \\[1em] \oint \vec{\text B} \cdot \vec{\text{dl}} = \text B \oint \text{dl}\end{aligned}$$

But $\oint \text{dl} = 2\text{πr}$, the circumference of the circle, and the current enclosed by the circle is I. Hence by Ampere's circuital law,

$$\begin{aligned}\text B(2\text{πr}) = \text µ_0 \text I \\[1em] \text B = \frac{\text µ_0}{2\text π}\frac{\text I}{\text r}\end{aligned}$$

The direction of $\vec{\text B} \space$is tangential to the circle, and is given by the right-hand thumb rule.

Ampere's circuital law : The line integral of the magnetic flux density $\vec{\text B} \space$around any closed path in free space is equal to µ₀ times the net current I threading through the area enclosed by the path,

$\oint \vec{\text B} \cdot \vec{\text{dl}} = \text µ_0 \text I$

Magnetic field of a long straight solenoid :

Consider a solenoid very long compared with its diameter, having n turns per unit length and carrying a current I₀. The field outside such a solenoid is very small compared with the field inside, and the lines of force inside are straight and parallel, showing a uniform field except near the edges.

Let abcd be a closed rectangular path in which the side ab of length x is parallel to the axis inside the solenoid, and the sides bc and da are very long so that cd is far from the solenoid where the field is negligibly small.

$\oint \vec{\text B} \cdot \vec{\text{dl}} = \int_\text a^\text b \vec{\text B} \cdot \vec{\text{dl}} + \int_\text b^\text c \vec{\text B} \cdot \vec{\text{dl}} + \int_\text c^\text d \vec{\text B} \cdot \vec{\text{dl}} + \int_\text d^\text a \vec{\text B} \cdot \vec{\text{dl}}$

Along bc and da the field is at right angles to $\vec{\text{dl}} \space$, so $\vec{\text B} \cdot \vec{\text{dl}} = \text B\ \text{dl} \cos 90° = 0$, and along cd the field itself is zero. Hence

$$\begin{aligned}\oint \vec{\text B} \cdot \vec{\text{dl}} = \int_\text a^\text b \text B\ \text{dl} \\[1em] \oint \vec{\text B} \cdot \vec{\text{dl}} = \text B \int_\text a^\text b \text{dl} \\[1em] \oint \vec{\text B} \cdot \vec{\text{dl}} = \text{Bx} \qquad \dots(\text i)\end{aligned}$$

The number of turns passing through the rectangle is nx, and the current in each turn is I₀, so the net current enclosed is

$\text I = \text{nx}\ \text I_0 \qquad \dots(\text{ii})$

Substituting (i) and (ii) in Ampere's circuital law,

$$\begin{aligned}\text{Bx} = \text µ_0 \text{nx}\ \text I_0 \\[1em] \text B = \text µ_0 \text n \text I_0\end{aligned}$$

Thus the field inside a long solenoid is independent of its length and diameter and is uniform over the cross-section. At the ends of the solenoid the field is one half of this value,

$\text B_{\text{end}} = \frac{1}{2}\text µ_0 \text n \text I_0$

Toroid : A toroid is a hollow circular ring on which a large number of turns of insulated conducting wire are closely wound. It may be viewed as a solenoid bent into a circular shape so as to close on itself.

Magnetic field inside the core of a toroid : Let the toroid have N turns in all and carry a current I₀. Let P be a point inside the core, and let a circle of radius a be drawn through P, lying within the space enclosed by the windings.

By symmetry the magnitude of $\vec{\text B} \space$is the same at all points of this circle and its direction is tangential at each point. Applying Ampere's circuital law to this circle,

$\oint \vec{\text B} \cdot \vec{\text{dl}} = \text µ_0 \text I \qquad \dots(\text i)$

Since $\vec{\text B} \space$and $\vec{\text{dl}} \space$are along the same direction at every element,

$$\begin{aligned}\oint \vec{\text B} \cdot \vec{\text{dl}} = \oint \text B\ \text{dl} \\[1em] \oint \vec{\text B} \cdot \vec{\text{dl}} = \text B(2\text{πa})\end{aligned}$$

The circle through P is threaded by all the N turns, each carrying the current I₀, so the net current enclosed is

$\text I = \text N \text I_0$

Substituting in equation (i),

$$\begin{aligned}\text B(2\text{πa}) = \text µ_0 \text N \text I_0 \\[1em] \text B = \frac{\text µ_0}{2\text π}\frac{\text N \text I_0}{\text a}\end{aligned}$$

If n is the number of turns per unit length of the toroid, then N = n(2πa) and

$\text B = \text µ_0 \text n \text I_0$

Difference from a tightly wound long solenoid : For a long solenoid B = µ₀nI₀ is the same at every point well inside, so the field is genuinely uniform. In a toroid the field

$\text B = \frac{\text µ_0}{2\text π}\frac{\text N \text I_0}{\text a}$

varies with the radius a of the circle drawn through the point, so the field within the core is not uniform; it is larger near the inner edge and smaller near the outer edge.

Further, the field of a toroid is confined entirely to its core. At a point such as Q in the empty space enclosed by the toroid the field is zero because no current is enclosed, and at a point R outside the toroid it is also zero because the circle through R encloses equal currents in opposite directions.

Defining vector equation of magnetic field : When a charge q moves with velocity $\vec{\text v} \space$in a magnetic field $\vec{\text B} \space$, the force acting upon it is

$\vec{\text F} = \text q(\vec{\text v} \times \vec{\text B})$

whose magnitude is

$$\begin{aligned}\text F = \text{Bqv} \sin \text θ \\[1em] \text B = \frac{\text F}{\text{qv} \sin \text θ}\end{aligned}$$

where θ is the angle between $\vec{\text v} \space$and $\vec{\text B} \space$. This relation defines the magnetic field (magnetic induction) $\vec{\text B} \space$.

SI unit in terms of base units : If F = 1 newton, q = 1 coulomb, v = 1 metre/second and θ = 90°, then

$$\begin{aligned}\text B = \frac{1\ \text N}{1\ \text C \times 1\ \text{m s}^{-1} \times 1} \\[1em] \text B = 1\ \frac{\text N}{\text A\ \text m} \\[1em] \text B = 1\ \text N \text A^{-1} \text m^{-1}\end{aligned}$$

since C s^-1 = A. This unit is called the tesla (T).

Expressing the newton in base units, N = kg m s^-2,

$$\begin{aligned}1\ \text T = \frac{\text{kg m s}^{-2}}{\text A\ \text m} \\[1em] 1\ \text T = \text{kg s}^{-2}\ \text A^{-1}\end{aligned}$$

Thus the SI unit of $\vec{\text B} \space$in terms of the fundamental units is kg s^-2 A^-1.

Motion of a charged particle in a uniform magnetic field : The path traced depends upon the angle θ between $\vec{\text v} \space$and $\vec{\text B} \space$.

(a) θ = 0 : The force F = Bqv sin 0° = 0, so the particle moves undeviated along a straight line.

(b) θ = 90° : The force F = Bqv is maximum and is always perpendicular to the velocity, so it does not change the speed of the particle but only the direction of motion. It therefore acts as the centripetal force and the particle describes a circle with a constant speed.

(c) θ neither 0 nor 90° : The velocity is resolved into v cos θ parallel to $\vec{\text B} \space$and v sin θ perpendicular to it. The parallel component gives a uniform linear motion and the perpendicular component a circular motion, so the resultant path is a helix whose axis is parallel to the field.

Radius of the circular path : For a particle moving perpendicular to the field, the magnetic force provides the centripetal force,

$$\begin{aligned}\text{qvB} = \frac{\text{mv}^2}{\text r} \\[1em] \text r = \frac{\text{mv}}{\text{qB}} \\[1em] \text r = \frac{\text p}{\text{qB}}\end{aligned}$$

Thus the radius of the path is directly proportional to the momentum (mv) of the particle. Dividing the numerator and the denominator by m,

$$\begin{aligned}\text r = \frac{\text v}{(\text q/\text m)\text B} \\[1em] \text r = \frac{\text v}{(\text e/\text m)\text B}\end{aligned}$$

so the radius is inversely proportional to the specific charge (e/m) of the particle.

Frequency of revolution : The particle traverses a distance 2πr in one revolution, so

$$\begin{aligned}\text T = \frac{2\text{πr}}{\text v} \\[1em] \text T = \frac{2\text π}{\text v} \times \frac{\text{mv}}{\text{qB}} \\[1em] \text T = \frac{2\text{πm}}{\text{qB}} \\[1em] \text n = \frac{1}{\text T} \\[1em] \text n = \frac{\text{qB}}{2\text{πm}}\end{aligned}$$

The expression for n contains only q, B and m, and not the speed v. Hence the frequency of revolution of the particle is independent of its speed. If the speed increases, the radius increases in the same proportion, so the time taken to complete one revolution remains unaltered.

Expression for the force on a current-carrying conductor :

The current in a conductor is the drift of free electrons through it. When the conductor is placed in a magnetic field, magnetic forces are imposed on these drifting electrons, and the resultant of these microscopic forces is the macroscopic force on the whole conductor.

Consider a portion of length l and cross-sectional area A of a conductor carrying a current I, placed in a magnetic field $\vec{\text B} \space$, the conductor making an angle θ with the field. Let n be the number of free electrons per unit volume and v_d the drift velocity of the electrons.

From the relation $\vec{\text F} = \text q(\vec{\text v} \times \vec{\text B})$, the magnitude of the force on each electron is

$\text F' = \text e\ \text v_\text d \text B \sin \text θ$

The number of electrons in the length l of the conductor is

$\text N = \text{nAl}$

Hence the total force on the length l of the conductor is

$$\begin{aligned}\text F = \text F' \times \text N \\[1em] \text F = (\text e\ \text v_\text d \text B \sin \text θ)(\text{nAl}) \\[1em] \text F = (\text{neA}\ \text v_\text d)\ \text B l \sin \text θ\end{aligned}$$

But neAv_d = I, the current in the conductor. Therefore

$\text F = \text{IB}l \sin \text θ$

and in vector form

$\vec{\text F} = \text I\ \vec{l} \times \vec{\text B}$

where $\vec{l} \space$is a displacement vector pointing along the conductor in the direction of the current.

Rule for the direction of the force : By Fleming's left-hand rule, if the fore-finger, the middle-finger and the thumb of the left hand are stretched at right angles mutually to one another, with the fore-finger along $\vec{\text B} \space$and the middle-finger in the direction of the current, then the thumb gives the direction of the force $\vec{\text F} \space$on the conductor. The same result is given by the right-hand palm rule no. 2.

Force between two parallel current-carrying conductors :

Let PQ and RS be two infinitely long, straight and parallel conductors placed in vacuum at a distance r apart, carrying currents I₁ and I₂ respectively in the same direction.

The magnitude of the magnetic field produced by the current I₁ at the position of the conductor RS is

$\text B_1 = \frac{\text µ_0}{2\text π}\frac{\text I_1}{\text r}$

By the right-hand palm rule no. 1, this field is perpendicular to the plane of the paper directed downward.

The conductor RS, carrying the current I₂, is thus situated in this field at right angles to it, so it experiences a magnetic force. For a length L of RS,

$$\begin{aligned}\text F = \text I_2 \text B_1 \text L \sin 90° \\[1em] \text F = \text I_2 \left(\frac{\text µ_0}{2\text π}\frac{\text I_1}{\text r}\right)\text L\end{aligned}$$

Therefore the force per metre length of RS is

$\frac{\text F}{\text L} = \frac{\text µ_0}{2\text π}\frac{\text I_1 \text I_2}{\text r}$

The same force per unit length acts on PQ due to the current I₂. If the directions of the currents are the same, the wires attract each other; if the directions are opposite, the wires repel each other.

Definition of 1 ampere : If the two currents are equal (I₁ = I₂ = I),

$\frac{\text F}{\text L} = \frac{\text µ_0}{2\text π}\frac{\text I^2}{\text r}$

Putting I = 1 A and r = 1 m,

$$\begin{aligned}\frac{\text F}{\text L} = \frac{4\text π \times 10^{-7}}{2\text π} \times \frac{1}{1} \\[1em] \frac{\text F}{\text L} = 2 \times 10^{-7}\ \text{N m}^{-1}\end{aligned}$$

Hence, one ampere is that steady current which, when maintained in each of two infinitely long, straight and parallel conductors of negligible cross-section placed in vacuum at a distance of 1 metre from each other, produces a force of 2 × 10^-7 newton per metre length on each conductor.

Cyclotron : The cyclotron was devised by Lawrence to accelerate positively-charged particles, such as protons and deuterons, to very high energies so that they could be used in disintegration experiments.

Principle : A charged particle gains kinetic energy equal to qV each time it is suitably subjected to an electric potential difference V. By repeating the process again and again, the particle can be made to acquire a very high energy. The magnetic field is used to bring the charged particle again and again to the situation of acceleration.

Construction : It consists of two horizontal D-shaped hollow metal segments D₁ and D₂, called the dees, with a small gap between them. An alternating potential difference of the order of 10⁵ volt, at a frequency of 10 to 15 megacycles/sec, is applied across the dees. An intense magnetic field $\vec{\text B} \space$is set up perpendicular to the plane of the dees by a large electromagnet. The whole space inside the dees is evacuated. An ion source is located at the centre S in the gap between the dees.

Working : Suppose an ion of mass m and charge +q emerges from the ion source at an instant when D₂ is at a negative potential. It is accelerated towards D₂ by the electric field in the gap and enters D₂ with a velocity v. Once inside, it is screened from the electric field by the metal walls of the dee, and under the action of the magnetic field it adopts a circular path of radius

$\text r = \frac{\text{mv}}{\text{qB}}$

The time required by the ion to complete a semi-circle is

$$\begin{aligned}\text t = \frac{\text{πr}}{\text v} \\[1em] \text t = \frac{\text{πm}}{\text{qB}}\end{aligned}$$

which shows that the time of passage of the ion through the dee is independent of the speed of the ion and of the radius of the circle; it depends only on the magnetic field B and the charge-to-mass ratio q/m of the ion.

The frequency of the applied potential difference is so adjusted that the ion emerges into the gap at the instant when the other dee is at a negative potential. It is therefore accelerated again while crossing the gap, and enters D₁ with a larger velocity, describing a semi-circle of greater radius, but taking exactly the same time. The process is repeated after every half-cycle, and the ion gains speed each time it passes from one dee to the other. Finally, on becoming energetic enough, it reaches the outer edge of a dee and is pulled out by a negatively-charged deflector plate.

Cyclotron frequency : The frequency ν₀ of the applied potential difference must be equal to the frequency ν of the circular revolution of the ion,

$$\begin{aligned}\text ν_0 = \text ν = \frac{1}{2\text t} \\[1em] \text ν_0 = \frac{\text{qB}}{2\text{πm}}\end{aligned}$$

This expression contains only q, B and m, and not the speed v of the particle. Hence the cyclotron frequency does not depend on the speed of the particles. If the speed increases, the radius of the path increases in the same proportion, so that the time of one revolution remains unchanged.

(i) The response of the student is incorrect.

The magnetic field of a long solenoid is B = µ₀nI, so B ∝ I. To make the field strength double, the current should be changed from I to 2I. The student increases the current by two times, that is from I to 3I, which would make the field three times the present value.

(ii) The response of the student is correct.

Soft iron is a ferromagnetic substance. When a rod of such a material is inserted inside the solenoid, the field becomes

$\text B = \text µ_0 \text µ_\text r \text n \text I$

and since µ_r is very large for soft iron, the field strength is greatly increased. Aluminium and copper are not ferromagnetic, so they would not produce this effect.

(iii) The response of the student is incorrect.

The field at the middle of the solenoid is B_inside = µ₀nI, while at one end only one half of the winding contributes, so

$\text B_{\text{end}} = \frac{\text µ_0}{2}\text n \text I$

Therefore

$\frac{\text B_{\text{inside}}}{\text B_{\text{end}}} = \frac{2}{1}$

that is, the ratio is 2 : 1 and not 1 : 1.

(a) The magnetic field is along the positive Y-axis and the particle moves along the positive X-axis, so

$$\begin{aligned}\vec{\text v} \times \vec{\text B} = \text{vB}(\hat{\text i} \times \hat{\text j}) \\[1em] \vec{\text v} \times \vec{\text B} = \text{vB}\ \hat{\text k}\end{aligned}$$

(i) For an electron, the charge is −e, so

$$\begin{aligned}\vec{\text F} = (-\text e)(\text{vB}\ \hat{\text k}) \\[1em] \vec{\text F} = -\text{evB}\ \hat{\text k}\end{aligned}$$

The Lorentz force is along the −Z-axis for an electron.

(ii) For a proton, the charge is +e, so

$$\begin{aligned}\vec{\text F} = (+\text e)(\text{vB}\ \hat{\text k}) \\[1em] \vec{\text F} = \text{evB}\ \hat{\text k}\end{aligned}$$

The Lorentz force is along the +Z-axis for a proton.

(b) The electron suffers no deflection, so the magnetic force on it is zero. Since F = Bqv sin θ and neither B nor v is zero, sin θ must be zero. Hence the direction of the magnetic field is parallel (or anti-parallel) to the direction of motion of the electron.

STEP 1 :

$\oint \vec{\text B} \cdot \vec{\text{dl}} = \text µ_0 (\text{Total current}) = \text µ_0 \text N \text I$

The circle drawn through the point inside the core is threaded by all the N turns of the toroid, each carrying the current I, so the total current enclosed is NI.

STEP 2 :

$\text B(2\text{πr}) = \text µ_0 \text N \text I$

Since $\vec{\text B} \space$and $\vec{\text{dl}} \space$are along the same direction at every point of the circle and B has the same magnitude at all these points, the line integral becomes B times the circumference 2πr of the circle.

STEP 3 :

$\text B = \text µ_0 \text n \text I$

Dividing by 2πr gives $\text B = \frac{\text µ_0 \text N \text I}{2\text{πr}}$, and writing $\text n = \frac{\text N}{2\text{πr}}$ for the number of turns per unit length of the toroid, this becomes B = µ₀nI.

The total force on the charge is the Lorentz force,

$\vec{\text F} = \text q(\vec{\text E} + \vec{\text v} \times \vec{\text B})$

(a) B = 0.

If E = 0 the electric force is already zero. The resultant force can be zero only if the magnetic force is also zero. As $\vec{\text v} \space$is perpendicular to $\vec{\text B} \space$here (θ = 90°), the magnetic force qvB sin 90° cannot vanish unless B = 0.

(b) E = 0.

The resultant force may be zero with B ≠ 0 provided the magnetic force itself vanishes, which happens when θ = 0° or 180°, that is, when the charge moves parallel or anti-parallel to $\vec{\text B} \space$. For the total force to be zero the electric force must then also be absent, so E = 0.

(c) B ≠ 0.

Here E ≠ 0, so an electric force $\vec{\text F}_\text e$ acts on the charge. It can be balanced only by an equal and opposite magnetic force $\vec{\text F}_\text m$, which requires a magnetic field, that is B ≠ 0. This is the condition qE = qvB, giving $\text v = \frac{\text E}{\text B}$.

(d) B = 0.

Here the velocity is not constant, so the balance qE = qvB cannot be maintained and the resultant force is not zero. This is the case when the electric field alone acts, that is B = 0; the charge is then continuously accelerated along $\vec{\text E} \space$and its velocity keeps changing.

(a) The resultant of the forces experienced by the arms QR and SP is zero.

The arms QR and SP are equally inclined to the wire AB and each of them has one half of its length nearer to AB and the other half farther from it. The magnitudes of the forces on QR and on SP are therefore the same, but the currents in these two arms are oppositely directed, so the two forces are opposite in direction and cancel each other.

(b) There is a force of attraction between AB and PQ, and a force of repulsion between AB and SR.

The current in PQ is parallel to the current in AB, and parallel currents attract. The current in SR is anti-parallel to the current in AB, and anti-parallel currents repel.

(c) The loop begins to move towards the wire AB.

The force per unit length between two parallel conductors is

$\frac{\text F}{\text L} = \frac{\text µ_0}{2\text π}\frac{\text I_1 \text I_2}{\text r}$

which is inversely proportional to the separation r. Since PQ is closer to AB than SR is, the force of attraction on PQ is stronger than the force of repulsion on SR. Hence there is a net attraction and the loop moves towards AB.

Biot-Savart's law : The magnetic field dB developed at a point due to a small current-element I dl is directly proportional to the current I, to the length dl of the element and to the sine of the angle θ between the element and the line joining it to the point, and inversely proportional to the square of the distance r between them,

$\text{dB} = \frac{\text µ_0}{4\text π}\frac{\text I\ \text{dl} \sin \text θ}{\text r^2}, \qquad \text d\vec{\text B} = \frac{\text µ_0}{4\text π}\frac{\text I(\vec{\text{dl}} \times \vec{\text r})}{\text r^3}$

(i) Magnetic field at a point on the axis of the coil :

Let a circular loop of radius a carry a current I, and let P be a point on its axis at a distance x from the centre O. Consider a small current-element of length dl at the top of the loop, at right angles to the plane of the page and directed outwards, and let r be its distance from P.

By Biot-Savart law, the field due to this element is

$\text{dB} = \frac{\text µ_0}{4\text π}\frac{\text I\ \text{dl} \sin \text θ}{\text r^2}$

Here the element is perpendicular to the line joining it to P, so θ = 90° and

$\text{dB} = \frac{\text µ_0}{4\text π}\frac{\text I\ \text{dl}}{\text r^2} \qquad \dots(\text i)$

The direction of $\text d\vec{\text B} \space$is in the plane of the page and at right angles to r. It is resolved into a component dB sin φ along the axis and a component dB cos φ at right angles to the axis.

Now consider the diametrically opposite element at the bottom of the loop, which is at right angles to the plane of the page but directed inwards. The field due to it is equal in magnitude, and its component at right angles to the axis is equal and opposite to that of the first element, so these components cancel, while the components along the axis are in the same direction and add up.

Imagining the whole loop to be divided into such current-elements, the resultant field at P is directed along the axis and its magnitude is

$\text B = \int \text{dB} \sin \text φ$

Putting the value of dB from equation (i) and $\sin \text φ = \frac{\text a}{\text r}$,

$$\begin{aligned}\text B = \int \frac{\text µ_0}{4\text π}\frac{\text I}{\text r^2} \times \frac{\text a}{\text r}\ \text{dl} \\[1em] \text B = \frac{\text µ_0}{4\text π}\frac{\text{Ia}}{\text r^3}\int \text{dl}\end{aligned}$$

since µ₀/4π, I and a are constants and r is the same for all the elements of the loop.

But $\int \text{dl} = 2\text{πa}$, the circumference of the loop, and $\text r = \sqrt{\text a^2 + \text x^2}$. Hence

$\text B = \frac{\text µ_0}{4\text π}\frac{2\text{πIa}^2}{(\text a^2 + \text x^2)^{3/2}}$

If the coil has N turns, each turn contributes equally, so

$\text B = \frac{\text µ_0}{4\text π}\frac{2\text{πNIa}^2}{(\text a^2 + \text x^2)^{3/2}}$

(ii) Magnetic field at the centre of the coil :

At the centre, x = 0, so

$$\begin{aligned}\text B = \frac{\text µ_0}{4\text π}\frac{2\text{πNI}}{\text a} \\[1em] \text B = \frac{\text µ_0 \text N \text I}{2\text a}\end{aligned}$$

Direction of the field :

The direction of $\vec{\text B} \space$is perpendicular to the plane of the coil, along its axis, and is given by the right-hand palm rule no. 1. When the current flows anticlockwise as seen by the observer, the field at the centre is directed towards the observer (out of the plane of the coil), and that face of the coil behaves as a north pole. When the current flows clockwise, the field is directed away from the observer (into the plane of the coil), and that face behaves as a south pole.

Ampere's circuital law : The line integral of the magnetic flux density $\vec{\text B} \space$around any closed path in free space is equal to µ₀ times the net current I threading through the area enclosed by the path,

$\oint \vec{\text B} \cdot \vec{\text{dl}} = \text µ_0 \text I$

Here I is the steady current that does not change with time, and the law is true for any assembly of currents and for any closed path.

Magnetic field due to an infinitely-long straight current-carrying wire :

Let a long straight wire carry a steady current I, and let P be a point at a perpendicular distance r from it.

Draw a circle of radius r through P, with its centre on the wire and its plane perpendicular to the wire. By symmetry, the magnitude of $\vec{\text B} \space$is the same at all points of this circle and its direction is tangential to the circle at every point. Choosing this circle as the closed path, $\vec{\text B} \space$and $\vec{\text{dl}} \space$are along the same direction, so

$$\begin{aligned}\oint \vec{\text B} \cdot \vec{\text{dl}} = \oint \text B\ \text{dl} \cos 0° \\[1em] \oint \vec{\text B} \cdot \vec{\text{dl}} = \text B \oint \text{dl} \\[1em] \oint \vec{\text B} \cdot \vec{\text{dl}} = \text B(2\text{πr})\end{aligned}$$

The current enclosed by the circle is I, so by Ampere's circuital law,

$$\begin{aligned}\text B(2\text{πr}) = \text µ_0 \text I \\[1em] \text B = \frac{\text µ_0}{2\text π}\frac{\text I}{\text r}\end{aligned}$$

Why the derivation is not valid for a wire of finite length :

The derivation rests on the symmetry of the field about the wire. Only for an infinitely long wire is the magnitude of $\vec{\text B} \space$the same at every point of the circle drawn through P, which allows B to be taken outside the integral as $\text B \oint \text{dl}$.

For a wire of finite length this symmetry no longer holds; the magnitude of $\vec{\text B} \space$differs from point to point on the circle, and the line integral cannot be evaluated so simply. The field of a finite conductor has to be obtained from Biot-Savart law instead, which gives

$\text B = \frac{\text µ_0}{4\text π}\frac{\text I}{\text r}(\sin \text φ_1 + \sin \text φ_2)$

Cyclotron :

Working principle : A charged particle gains a kinetic energy qV every time it is subjected to a potential difference V. If the particle is made to face such a potential difference again and again, it can be accelerated to a very high energy. In a cyclotron a strong magnetic field, applied perpendicular to the plane of the dees, makes the particle travel along semi-circular paths and brings it back to the gap again and again, where an alternating potential difference accelerates it each time.

Inside a dee the particle moves along a semi-circle of radius

$\text r = \frac{\text{mv}}{\text{qB}}$

and the time taken to complete it is

$$\begin{aligned}\text t = \frac{\text{πr}}{\text v} \\[1em] \text t = \frac{\text{πm}}{\text{qB}}\end{aligned}$$

which is independent of the speed of the particle. The frequency of the applied potential difference must therefore satisfy the resonance condition

$\text ν_0 = \frac{\text{qB}}{2\text{πm}}$

The radius of the path grows at each crossing, but the time of passage through a dee stays the same, so the particle always arrives at the gap in step with the alternating potential difference. It is finally pulled out at the outer edge of a dee by a negatively-charged deflector plate, with a kinetic energy

$$\begin{aligned}\text K = \frac{1}{2}\text m\text v_\text{max}^2 \\[1em] \text K = \frac{\text q^2 \text B^2 \text R^2}{2\text m}\end{aligned}$$

where R is the outside radius of the dees.

Two applications :

(i) It is used to accelerate positively-charged particles, such as protons and deuterons, to very high energies for bringing about artificial transmutation of nuclei.

(ii) The high-energy particles from a cyclotron are used in nuclear disintegration experiments.

Limitation in accelerating a light particle such as an electron : The cyclotron cannot accelerate particles to velocities as high as that comparable with the velocity of light, because at such velocities the mass of the particle increases with increasing velocity according to

$\text m = \frac{\text m_0}{\sqrt{1 - \frac{\text v^2}{\text c^2}}}$

where m₀ is the rest mass. As m increases, the time $\text t = \frac{\text{πm}}{\text{qB}}$ taken to complete a semi-circle also increases, so the particle gets more and more out of step with the applied alternating potential difference and the resonance condition fails.

For a given energy the velocity of an electron is much larger than that of a proton or a deuteron, so this relativistic increase of mass becomes important very quickly. Hence electrons get out of step with the applied potential difference almost at once, and a cyclotron cannot be used to accelerate electrons to high energies.

(i) The electrons move towards the right in region II and the path bends downwards, so the force on them is directed towards the centre of the circular arc.

The electron is negatively charged, so the direction of current is taken opposite to its direction of motion. Applying Fleming's left-hand rule with the current in this direction and the force towards the centre of the arc, the magnetic field must be directed downwards into the plane of the paper, that is, perpendicular to the plane of the paper.

(ii) The solenoid current controls the magnetic field, and for a solenoid B = µ₀nI_s, so B ∝ I_s. The radius of the path of the electrons is

$\text r = \frac{\text{mv}}{\text{qB}} \quad \Rightarrow \quad \text r \propto \frac{1}{\text B} \propto \frac{1}{\text I_\text s}$

On increasing I_s above its previous value, the radius of the path decreases, so the electrons bend more sharply. Hence the electrons are expected to follow path 3.

(iii) Since $\text r \propto \frac{1}{\text I_\text s}$, halving the solenoid current doubles the radius. Hence the radius of the path becomes 2r.

(iv) For a long solenoid,

$$\begin{aligned}\text B = \text µ_0 \text n \text I \\[1em] \text B = \text µ_0 \frac{\text N}{l}\text I \\[1em] \text B \propto \text N\end{aligned}$$

The field does not depend on the area of cross-section of the solenoid. Hence, on making the number of turns double while the current remains the same, the magnitude of the magnetic field inside the solenoid becomes double.

(v) The electrons are accelerated through a potential difference V, so

$$\begin{aligned}\frac{1}{2}\text{mv}^2 = \text{qV} \\[1em] \text v = \sqrt{\frac{2\text{qV}}{\text m}} \\[1em] \text r = \frac{\text{mv}}{\text{qB}} \\[1em] \text r = \frac{\sqrt{2\text{mqV}}}{\text{qB}} \\[1em] \text r \propto \text V^{1/2}\end{aligned}$$

Hence, n = $\frac{1}{2}$.

(i) When the charge particle passes undeflected, the electric force balances the magnetic force,

$$\begin{aligned}\text{qE} = \text{qvB} \\[1em] \text v = \frac{\text E}{\text B}\end{aligned}$$

Hence, the charge particle moves undeflected when its velocity is equal to $\frac{\text E}{\text B}$.

(ii) The magnetic force on a particle entering perpendicular to the field is

$\text F = \text{qvB}$

Since all the particles enter with the same velocity in the same field, the force depends only on the magnitude of the charge. The neutron is uncharged and experiences no force; the proton and the electron carry a charge of magnitude e; the α-particle carries a charge 2e, which is the largest here.

Hence, the α-particle will experience the maximum force.

(iii) All of these.

The particle moves with a constant velocity, so the net force on it is zero. This is possible in three ways : both the fields may be absent (E = 0, B = 0); both the fields may be present with the electric and the magnetic forces equal and opposite (E ≠ 0, B ≠ 0); or the electric field may be absent while the particle moves parallel to the magnetic field (E = 0, B ≠ 0). Hence all the three options are possible.

(iv) For a particle entering at right angles to the field with kinetic energy K,

$\text r = \frac{\sqrt{2\text{mK}}}{\text{qB}} \quad \Rightarrow \quad \text r \propto \sqrt{\text m}$

since K, q and B are the same for all the three particles. The deuteron (₁H²) has the greatest mass among the proton, the electron and the deuteron.

Hence, the deuteron follows the curved path of larger radius.

(v) With the magnetic field switched off, only the electric force qE acts on the particle. This force is constant in magnitude and direction, and is perpendicular to the initial velocity of the particle.

Hence, the charge particle will follow a parabolic path.

Given,

• I = 50 A, flowing from north to south
• r = 2.5 m

For a long straight current-carrying conductor,

$$\begin{aligned}\text B = \frac{\text μ_0}{2\text π}\frac{\text I}{\text r} \\[1em] \text B = \frac{\text μ_0}{4\text π}\frac{2\text I}{\text r}\end{aligned}$$

Substituting the values,

$$\begin{aligned}\text B = (10^{-7}) \times \frac{2 \times 50}{2.5} \\[1em] \text B = \frac{10^{-5}}{2.5} \\[1em] \text B = 4.0 \times 10^{-6}\ \text T\end{aligned}$$

By the right-hand thumb rule, with the current directed from north to south, the field at a point east of the wire is directed vertically upwards.

Hence, the magnetic field at the point is 4.0 × 10^-6 T, directed vertically upwards.

Given,

• I₁ = 2 A, at a distance r₁ = 4 cm = 0.04 m from P
• I₂ = 1.5 A, at a distance r₂ = 3 cm = 0.03 m from P

(i) The field at P due to each wire is $\text B = \frac{\text μ_0}{4\text π}\frac{2\text I}{\text r}$.

Due to the current I₁,

$$\begin{aligned}\text B_1 = (10^{-7}) \times \frac{2 \times 2}{0.04} \\[1em] \text B_1 = 1 \times 10^{-5}\ \text T\end{aligned}$$

Due to the current I₂,

$$\begin{aligned}\text B_2 = (10^{-7}) \times \frac{2 \times 1.5}{0.03} \\[1em] \text B_2 = 1 \times 10^{-5}\ \text T\end{aligned}$$

Applying the right-hand thumb rule to each wire, both the fields at P are perpendicular to the plane of the paper and directed downwards, so they add up,

$$\begin{aligned}\text B = \text B_1 + \text B_2 \\[1em] \text B = 2 \times 10^{-5}\ \text T\end{aligned}$$

Hence, the magnetic field at P is 2 × 10^-5 T, perpendicular to the plane of paper pointing downwards.

(ii) If the direction of the current in one of the wires is reversed, the field due to that wire at P is also reversed. The two fields are then equal in magnitude but opposite in direction,

$$\begin{aligned}\text B = \text B_1 - \text B_2 \\[1em] \text B = 0\end{aligned}$$

Hence, the magnetic field at P would then be zero.

Given,

• I = 5 microampere = 5 × 10^-6 A
• r = 50 cm = 0.5 m

(i) If n electrons pass through a point per second, the current is

$$\begin{aligned}\text I = \text{ne} \quad \Rightarrow \quad \text n = \frac{\text I}{\text e} \\[1em] \text n = \frac{5 \times 10^{-6}}{1.6 \times 10^{-19}} \\[1em] \text n = 3.125 \times 10^{13}\end{aligned}$$

Hence, 3.125 × 10¹³ electrons pass through a point per second.

(ii) The magnetic field produced by the beam at a distance r is

$$\begin{aligned}\text B = \frac{\text μ_0}{4\text π}\frac{2\text I}{\text r} \\[1em] \text B = (10^{-7}) \times \frac{2 \times 5 \times 10^{-6}}{0.5} \\[1em] \text B = 2.0 \times 10^{-12}\ \text T\end{aligned}$$

Hence, the magnetic field produced at a distance of 50 cm is 2.0 × 10^-12 T.

Given,

• I₁ = I₂ = 4 A
• separation = 16 cm, so the mid-point is at r = 8 cm = 0.08 m from each wire

The field at the mid-point due to each wire is

$$\begin{aligned}\text B_1 = \text B_2 = \frac{\text μ_0}{4\text π}\frac{2\text I}{\text r} \\[1em] \text B_1 = \text B_2 = (10^{-7}) \times \frac{2 \times 4}{0.08} \\[1em] \text B_1 = \text B_2 = 1 \times 10^{-5}\ \text T\end{aligned}$$

(i) Currents in the same direction : At the mid-point the two fields are equal in magnitude but opposite in direction, so

$$\begin{aligned}\text B = \text B_1 - \text B_2 \\[1em] \text B = 0\end{aligned}$$

Hence, the field at the mid-point is zero.

(ii) Currents in opposite directions : The two fields are now in the same direction, so

$$\begin{aligned}\text B = \text B_1 + \text B_2 \\[1em] \text B = 2 \times 10^{-5}\ \text T\end{aligned}$$

Hence, the field at the mid-point is 2 × 10^-5 T.

Given,

• N = 100 turns
• a = 8.0 cm = 0.08 m
• I = 0.40 A

The magnetic field at the centre of a circular coil is

$\text B = \frac{\text μ_0 \text N \text I}{2\text a}$

Substituting the values,

$$\begin{aligned}\text B = \frac{(4\text π \times 10^{-7})(100)(0.40)}{2 \times 0.08} \\[1em] \text B = \frac{4\text π \times 10^{-7} \times 40}{0.16} \\[1em] \text B = \text π \times 10^{-4} \\[1em] \text B = 3.14 \times 10^{-4}\ \text T\end{aligned}$$

Hence, the magnetic field at the centre of the coil is 3.14 × 10^-4 T.

Given,

• charge, q = 2e = 2 × 1.6 × 10^-19 = 3.2 × 10^-19 C
• a = 1.0 Å = 1.0 × 10^-10 m
• v = 2.0 × 10⁶ ms^-1

The revolving α-particle is equivalent to a current loop. The time taken for one revolution is $\text T = \frac{2\text{πa}}{\text v}$, so the equivalent current is

$$\begin{aligned}\text I = \frac{\text q}{\text T} \\[1em] \text I = \frac{\text{qv}}{2\text{πa}} \\[1em] \text I = \frac{(3.2 \times 10^{-19})(2.0 \times 10^6)}{2 \times 3.14 \times 10^{-10}} \\[1em] \text I = \frac{6.4 \times 10^{-13}}{6.28 \times 10^{-10}} \\[1em] \text I = 1.019 \times 10^{-3}\ \text A\end{aligned}$$

The magnetic field at the centre of the circular path is

$$\begin{aligned}\text B = \frac{\text μ_0 \text I}{2\text a} \\[1em] \text B = \frac{(4\text π \times 10^{-7})(1.019 \times 10^{-3})}{2 \times 10^{-10}} \\[1em] \text B = 6.4\ \text N \text A^{-1} \text m^{-1}\end{aligned}$$

Hence, the magnetic field produced at the centre of the circular path is 6.4 N A^-1 m^-1.

Given,

• r = 5.1 × 10^-11 m
• frequency, ν = 6.8 × 10¹⁵ s^-1

The revolving electron is equivalent to a current loop, and the equivalent current is

$$\begin{aligned}\text I = \text{eν} \\[1em] \text I = (1.6 \times 10^{-19})(6.8 \times 10^{15}) \\[1em] \text I = 1.088 \times 10^{-3}\ \text A\end{aligned}$$

The magnetic field at the centre of the orbit is

$$\begin{aligned}\text B = \frac{\text μ_0 \text I}{2\text r} \\[1em] \text B = \frac{(4\text π \times 10^{-7})(1.088 \times 10^{-3})}{2 \times 5.1 \times 10^{-11}} \\[1em] \text B = \frac{1.367 \times 10^{-9}}{1.02 \times 10^{-10}} \\[1em] \text B = 13.4\ \text N \text A^{-1} \text m^{-1}\end{aligned}$$

Hence, the magnetic field set up at the centre of the orbit is 13.4 N A^-1 m^-1.

Given,

• R = 10 cm = 0.1 m
• I = 8 A

The field at C is the resultant of the field due to the long straight portion of the wire and that due to the circular loop.

Field due to the straight wire :

$$\begin{aligned}\text B_1 = \frac{\text μ_0}{4\text π}\frac{2\text I}{\text R} \\[1em] \text B_1 = (10^{-7}) \times \frac{2 \times 8}{0.1} \\[1em] \text B_1 = 1.6 \times 10^{-5}\ \text T\end{aligned}$$

By the right-hand thumb rule this field is perpendicular to the plane of the paper, pointing downwards.

Field due to the loop :

$$\begin{aligned}\text B_2 = \frac{\text μ_0 \text I}{2\text R} \\[1em] \text B_2 = \frac{(4\text π \times 10^{-7})(8)}{2 \times 0.1} \\[1em] \text B_2 = 5.03 \times 10^{-5}\ \text T\end{aligned}$$

This field is perpendicular to the plane of the paper, pointing upwards, that is, just opposite to B₁.

Resultant field :

$$\begin{aligned}\text B = \text B_2 - \text B_1 \\[1em] \text B = (5.03 - 1.6) \times 10^{-5} \\[1em] \text B = 3.4 \times 10^{-5}\ \text T\end{aligned}$$

Hence, the magnetic field at the centre C is 3.4 × 10^-5 T, perpendicular to the plane of paper pointing upwards.

Given, two semi-circular current-loops of radii a₁ and a₂ having the common centre O, each carrying a current I.

The magnetic field at the centre due to a circular arc subtending an angle θ (in radians) at the centre is

$\text B_{\text{arc}} = \left(\frac{\text μ_0}{4\text π}\frac{\text I}{\text a}\right)\text θ$

For a semi-circle, θ = π, so each loop gives

$\text B_1 = \frac{\text μ_0 \text I}{4\text a_1}, \qquad \text B_2 = \frac{\text μ_0 \text I}{4\text a_2}$

The current flows in the same sense in both the semi-circles, so by the right-hand palm rule no. 1 both the fields at O are perpendicular to the plane of the paper, directed downwards, and they add up,

$$\begin{aligned}\text B = \frac{\text μ_0 \text I}{4\text a_1} + \frac{\text μ_0 \text I}{4\text a_2} \\[1em] \text B = \frac{\text μ_0 \text I}{4}\left(\frac{1}{\text a_1} + \frac{1}{\text a_2}\right)\end{aligned}$$

Hence, the magnetic field at the common centre O is $\frac{\text μ_0 \text I}{4}\left(\frac{1}{\text a_1} + \frac{1}{\text a_2}\right)$, perpendicular to the plane of paper, directed downwards.

Given, a bent wire carrying a current I, consisting of an arc of radius a₁ subtending 90° at O and an arc of radius a₂ subtending 270° at O.

The field at the centre due to a circular arc subtending an angle θ (in radians) is

$\text B_{\text{arc}} = \left(\frac{\text μ_0}{4\text π}\frac{\text I}{\text a}\right)\text θ$

For the arc of radius a₁, $\text θ_1 = 90° = \frac{\text π}{2}$,

$$\begin{aligned}\text B_1 = \frac{\text μ_0}{4\text π}\frac{\text I}{\text a_1} \times \frac{\text π}{2} \\[1em] \text B_1 = \frac{\text μ_0 \text I}{8\text a_1}\end{aligned}$$

For the arc of radius a₂, $\text θ_2 = 270° = \frac{3\text π}{2}$,

$$\begin{aligned}\text B_2 = \frac{\text μ_0}{4\text π}\frac{\text I}{\text a_2} \times \frac{3\text π}{2} \\[1em] \text B_2 = \frac{3\text μ_0 \text I}{8\text a_2}\end{aligned}$$

The straight radial portions pass through O, so for them θ = 0 and they develop no field at O. Both the arc fields are in the same direction, so

$\text B = \frac{\text μ_0 \text I}{8}\left(\frac{1}{\text a_1} + \frac{3}{\text a_2}\right)$

Hence, the magnetic field at O is $\frac{\text μ_0 \text I}{8}\left(\frac{1}{\text a_1} + \frac{3}{\text a_2}\right)$, perpendicular to the page, directed downwards.

Given, two circular arcs of radii a₁ and a₂, each subtending an angle α at the centre O, joined by radial straight portions and carrying a current I.

The field at the centre due to an arc subtending an angle α at the centre is

$\text B_{\text{arc}} = \left(\frac{\text μ_0}{4\text π}\frac{\text I}{\text a}\right)\text α$

So the two arcs give

$\text B_1 = \frac{\text μ_0 \text{Iα}}{4\text π \text a_1}, \qquad \text B_2 = \frac{\text μ_0 \text{Iα}}{4\text π \text a_2}$

The straight portions lie along the radii through O, so for them θ = 0 and they develop no field at O.

The current traverses the inner arc and the outer arc in opposite senses, so the two fields at O are oppositely directed and

$$\begin{aligned}\text B = \text B_1 - \text B_2 \\[1em] \text B = \frac{\text μ_0 \text{Iα}}{4\text π}\left(\frac{1}{\text a_1} - \frac{1}{\text a_2}\right)\end{aligned}$$

Hence, the magnetic field at O is $\frac{\text μ_0 \text{Iα}}{4\text π}\left(\frac{1}{\text a_1} - \frac{1}{\text a_2}\right)$, perpendicular to the plane of paper, directed upwards.

Given,

• wire A : I₁ = 2 A, R₁ = 3 cm = 0.03 m, the arc subtending 360° − 90° = 270° at O
• wire B : I₂ = 3 A, R₂ = 5 cm = 0.05 m, the arc subtending 360° − 60° = 300° at O

The field at the centre due to an arc subtending an angle θ (in radians) is

$\text B_{\text{arc}} = \left(\frac{\text μ_0}{4\text π}\frac{\text I}{\text R}\right)\text θ$

Due to wire A, $\text θ_1 = 270° = \frac{3\text π}{2}$ rad,

$$\begin{aligned}\text B_1 = (10^{-7}) \times \frac{2}{0.03} \times \frac{3\text π}{2} \\[1em] \text B_1 = (10^{-7})(66.67)(4.712) \\[1em] \text B_1 = 3.14 \times 10^{-5}\ \text T\end{aligned}$$

Due to wire B, $\text θ_2 = 300° = \frac{5\text π}{3}$ rad,

$$\begin{aligned}\text B_2 = (10^{-7}) \times \frac{3}{0.05} \times \frac{5\text π}{3} \\[1em] \text B_2 = (10^{-7})(60)(5.236) \\[1em] \text B_2 = 3.14 \times 10^{-5}\ \text T\end{aligned}$$

Both the arcs carry their currents in the same sense, so the two fields at O are in the same direction and add up,

$$\begin{aligned}\text B = \text B_1 + \text B_2 \\[1em] \text B = 3.14 \times 10^{-5} + 3.14 \times 10^{-5} \\[1em] \text B = 6.28 \times 10^{-5}\ \text T\end{aligned}$$

Hence, the net magnetic field at the common centre O is 6.28 × 10^-5 T.

Given, a conductor carrying a current i bent into concentric semicircles of radii a, 2a, 4a, 8a, ... , the trend continuing up to infinity.

The field at the centre due to a semicircle of radius a is

$\text B = \frac{\text μ_0 \text i}{4\text a}$

The successive semicircles are traversed in opposite senses, so their fields at O are alternately opposite in direction. Hence

$$\begin{aligned}\text B = \frac{\text μ_0 \text i}{4\text a} - \frac{\text μ_0 \text i}{4(2\text a)} + \frac{\text μ_0 \text i}{4(4\text a)} - \frac{\text μ_0 \text i}{4(8\text a)} + \dots \\[1em] \text B = \frac{\text μ_0 \text i}{4\text a}\left[1 - \frac{1}{2} + \frac{1}{4} - \frac{1}{8} + \dots\right]\end{aligned}$$

The series in the bracket is an infinite geometric series with first term 1 and common ratio $\text r = -\frac{1}{2}$, so its sum is

$$\begin{aligned}\text S = \frac{1}{1 - \left(-\frac{1}{2}\right)} \\[1em] \text S = \frac{1}{\frac{3}{2}} \\[1em] \text S = \frac{2}{3} \\[1em] \text B = \frac{\text μ_0 \text i}{4\text a} \times \frac{2}{3} \\[1em] \text B = \frac{\text μ_0 \text i}{6\text a}\end{aligned}$$

Hence, the magnetic field at the centre O is $\frac{\text μ_0 \text i}{6\text a}$.

Given,

• i = 2.5 A
• the circle x² + y² = 9 cm², so a = 3 cm = 0.03 m
• the point (0, 0, 4 cm), so x = 4 cm = 0.04 m on the axis

The field at a point on the axis of a circular loop is

$\text B = \frac{\text μ_0}{4\text π}\frac{2\text{πia}^2}{(\text a^2 + \text x^2)^{3/2}}$

Now,

$$\begin{aligned}\text a^2 + \text x^2 = (0.03)^2 + (0.04)^2 \\[1em] \text a^2 + \text x^2 = 0.0009 + 0.0016 \\[1em] \text a^2 + \text x^2 = 0.0025 \\[1em] (\text a^2 + \text x^2)^{3/2} = (0.0025)^{3/2} \\[1em] (\text a^2 + \text x^2)^{3/2} = 1.25 \times 10^{-4}\end{aligned}$$

Substituting,

$$\begin{aligned}\text B = (10^{-7}) \times \frac{2 \times 3.14 \times 2.5 \times (0.03)^2}{1.25 \times 10^{-4}} \\[1em] \text B = (10^{-7}) \times \frac{1.413 \times 10^{-2}}{1.25 \times 10^{-4}} \\[1em] \text B = 36\text π \times 10^{-7}\ \text T\end{aligned}$$

The point lies on the positive Z-axis, and by the right-hand palm rule no. 1 the field is directed along the +Z-axis.

Hence, the magnetic field at the point (0, 0, 4 cm) is 36π × 10^-7 T, directed along the +ve Z-axis.

Given,

• N = 500 turns
• length, l = 40 cm = 0.4 m
• I = 1.0 A

The number of turns per metre length is

$$\begin{aligned}\text n = \frac{\text N}{l} \\[1em] \text n = \frac{500}{0.4} \\[1em] \text n = 1250\ \text{turns m}^{-1}\end{aligned}$$

For a long solenoid, the field at the axis inside is

$$\begin{aligned}\text B = \text μ_0 \text n \text I \\[1em] \text B = (4\text π \times 10^{-7})(1250)(1.0) \\[1em] \text B = 1.57 \times 10^{-3}\ \text N \text A^{-1} \text m^{-1}\end{aligned}$$

Hence, the magnetic field at the axis inside the solenoid is 1.57 × 10^-3 N A^-1 m^-1.

Given,

• length, l = 80 cm = 0.8 m
• 5 layers of windings of 400 turns each, so N = 5 × 400 = 2000 turns
• I = 8.0 A

The number of turns per metre length is

$$\begin{aligned}\text n = \frac{\text N}{l} \\[1em] \text n = \frac{2000}{0.8} \\[1em] \text n = 2500\ \text{turns m}^{-1}\end{aligned}$$

Near the centre of a long solenoid,

$$\begin{aligned}\text B = \text μ_0 \text n \text I \\[1em] \text B = (4\text π \times 10^{-7})(2500)(8.0) \\[1em] \text B = 2.5 \times 10^{-2}\ \text T\end{aligned}$$

The field is independent of the diameter of the solenoid, so the value 1.8 cm is not needed.

Hence, the magnetic field inside the solenoid near the centre is 2.5 × 10^-2 T.

Given,

• B = 4.0 × 10^-2 T
• length, l = 50 cm = 0.5 m
• I = 8.0 A, π = 3.14

For a long solenoid, B = μ₀nI, so the number of turns per metre is

$$\begin{aligned}\text n = \frac{\text B}{\text μ_0 \text I} \\[1em] \text n = \frac{4.0 \times 10^{-2}}{(4 \times 3.14 \times 10^{-7})(8.0)} \\[1em] \text n = \frac{4.0 \times 10^{-2}}{1.0048 \times 10^{-5}} \\[1em] \text n = 3981\ \text{turns m}^{-1}\end{aligned}$$

The total number of turns in the solenoid is

$$\begin{aligned}\text N = \text n \times l \\[1em] \text N = 3981 \times 0.5 \approx 1990\end{aligned}$$

Hence, the number of turns in the solenoid is 1990.

Given,

• v = 5 × 10⁷ m s^-1
• B = 1 Wb m^-2
• θ = 30°
• q = e = 1.6 × 10^-19 C

The force on a charge moving in a magnetic field is

$\text F = \text{Bqv} \sin \text θ$

Substituting the values,

$$\begin{aligned}\text F = (1)(1.6 \times 10^{-19})(5 \times 10^7)(\sin 30°) \\[1em] \text F = (1.6 \times 10^{-19})(5 \times 10^7)(0.5) \\[1em] \text F = 4.0 \times 10^{-12}\ \text N\end{aligned}$$

Hence, the force on the electron is 4.0 × 10^-12 N.

Given,

• v = 2.0 × 10⁸ m s^-1, vertically upwards
• B = 0.50 N A^-1 m^-1, towards west
• m_e = 9.1 × 10^-31 kg

The velocity is perpendicular to the field, so θ = 90° and

$$\begin{aligned}\text F = \text{Bev} \\[1em] \text F = (0.50)(1.6 \times 10^{-19})(2.0 \times 10^8) \\[1em] \text F = 1.6 \times 10^{-11}\ \text N\end{aligned}$$

The electron is negatively charged, so the direction of current is vertically downwards. Applying Fleming's left-hand rule with the current downwards and the field towards the west, the force on the electron is directed towards the north.

The acceleration of the electron is

$$\begin{aligned}\text a = \frac{\text F}{\text m} \\[1em] \text a = \frac{1.6 \times 10^{-11}}{9.1 \times 10^{-31}} \\[1em] \text a = 1.8 \times 10^{19}\ \text m\ \text s^{-2}\end{aligned}$$

Hence, the force on the electron is 1.6 × 10^-11 N towards north, and its acceleration is 1.8 × 10¹⁹ m s^-2.

Given,

• v = 2.0 × 10⁷ ms^-1, towards east
• B = 0.34 T, towards north
• m_p = 1.7 × 10^-27 kg, q = 1.6 × 10^-19 C

(i) The velocity is perpendicular to the field, so

$$\begin{aligned}\text F = \text{qvB} \\[1em] \text F = (1.6 \times 10^{-19})(2.0 \times 10^7)(0.34) \\[1em] \text F = 1.09 \times 10^{-12}\ \text N\end{aligned}$$

Taking east along $\hat{\text i}$ and north along $\hat{\text j}$, $\vec{\text v} \times \vec{\text B} = \text{vB}(\hat{\text i} \times \hat{\text j}) = \text{vB}\ \hat{\text k}$, so the force on the proton is directed vertically upwards.

(ii) The radius of the proton's path is

$$\begin{aligned}\text r = \frac{\text{mv}}{\text{qB}} \\[1em] \text r = \frac{(1.7 \times 10^{-27})(2.0 \times 10^7)}{(1.6 \times 10^{-19})(0.34)} \\[1em] \text r = \frac{3.4 \times 10^{-20}}{5.44 \times 10^{-20}} \\[1em] \text r = 0.625\ \text m\end{aligned}$$

(iii) Let the lateral displacement be y while the proton advances a distance x = 0.20 m along the east. From the geometry of the circular path,

$$\begin{aligned}\text y = \text r - \sqrt{\text r^2 - \text x^2} \\[1em] \text y = 0.625 - \sqrt{(0.625)^2 - (0.20)^2} \\[1em] \text y = 0.625 - \sqrt{0.390625 - 0.04} \\[1em] \text y = 0.625 - \sqrt{0.350625} \\[1em] \text y = 0.625 - 0.592 \\[1em] \text y = 0.033\ \text m\end{aligned}$$

Hence, the force is 1.09 × 10^-12 N vertically upwards, the radius of the path is 0.625 m and the lateral displacement is 0.033 m.

Note: The textbook answer 0.032 m (approx.) results from the approximation y ≈ x²/2r; the exact value is 0.033 m.

Given,

• kinetic energy, K = 2 MeV = 2 × 10⁶ × 1.6 × 10^-19 = 3.2 × 10^-13 J
• B = 2.5 T
• m_p = 1.65 × 10^-27 kg, q = 1.6 × 10^-19 C

The speed of the proton is obtained from $\text K = \frac{1}{2}\text{mv}^2$,

$$\begin{aligned}\text v = \sqrt{\frac{2\text K}{\text m}} \\[1em] \text v = \sqrt{\frac{2 \times 3.2 \times 10^{-13}}{1.65 \times 10^{-27}}} \\[1em] \text v = \sqrt{3.879 \times 10^{14}} \\[1em] \text v = 1.97 \times 10^7\ \text m\ \text s^{-1}\end{aligned}$$

The proton moves perpendicular to the field, so

$$\begin{aligned}\text F = \text{qvB} \\[1em] \text F = (1.6 \times 10^{-19})(1.97 \times 10^7)(2.5) \\[1em] \text F = 7.88 \times 10^{-12}\ \text N\end{aligned}$$

Hence, the force on the proton is 7.88 × 10^-12 N.

Given,

• I = 3 A in the conductor PQ
• v = 2 × 10⁷ ms^-1, parallel to the conductor but opposite to the current (downwards)
• r = 0.6 m

The magnetic field produced by the conductor at the position of the proton is

$$\begin{aligned}\text B = \frac{\text μ_0}{4\text π}\frac{2\text I}{\text r} \\[1em] \text B = (10^{-7}) \times \frac{2 \times 3}{0.6} \\[1em] \text B = 1 \times 10^{-6}\ \text T\end{aligned}$$

This field is perpendicular to the plane of the paper, directed into it (right-hand thumb rule), so the proton moves at right angles to it and

$$\begin{aligned}\text F = \text{qvB} \\[1em] \text F = (1.6 \times 10^{-19})(2 \times 10^7)(1 \times 10^{-6}) \\[1em] \text F = 3.2 \times 10^{-18}\ \text N\end{aligned}$$

Applying Fleming's left-hand rule, this force on the proton is directed towards the right, that is, away from the wire.

Hence, the force acting on the proton is 3.2 × 10^-18 N, towards right away from the wire.

Given,

• accelerating potential, V = 300 V
• distance from the wire, r = 4 mm = 4 × 10^-3 m
• I = 5 A

The speed acquired by the electron is obtained from $\frac{1}{2}\text{mv}^2 = \text{eV}$,

$$\begin{aligned}\text v = \sqrt{\frac{2\text{eV}}{\text m}} \\[1em] \text v = \sqrt{\frac{2(1.6 \times 10^{-19})(300)}{9.1 \times 10^{-31}}} \\[1em] \text v = \sqrt{1.055 \times 10^{14}} \\[1em] \text v = 1.03 \times 10^7\ \text m\ \text s^{-1}\end{aligned}$$

The magnetic field of the wire at the position of the electron is

$$\begin{aligned}\text B = \frac{\text μ_0}{4\text π}\frac{2\text I}{\text r} \\[1em] \text B = (10^{-7}) \times \frac{2 \times 5}{4 \times 10^{-3}} \\[1em] \text B = 2.5 \times 10^{-4}\ \text T\end{aligned}$$

Taking the electron to move parallel to the wire, its velocity is perpendicular to this field, so

$$\begin{aligned}\text F = \text{evB} \\[1em] \text F = (1.6 \times 10^{-19})(1.03 \times 10^7)(2.5 \times 10^{-4}) \\[1em] \text F = 4 \times 10^{-16}\ \text N\end{aligned}$$

Hence, the force acting on the electron is 4 × 10^-16 N.

Given,

• B = 6.5 G = 6.5 × 10^-4 T
• v = 4.8 × 10⁶ ms^-1, normal to the field
• m_e = 9.1 × 10^-31 kg, e = 1.6 × 10^-19 C

Radius of the circular orbit :

$$\begin{aligned}\text r = \frac{\text{mv}}{\text{eB}} \\[1em] \text r = \frac{(9.1 \times 10^{-31})(4.8 \times 10^6)}{(1.6 \times 10^{-19})(6.5 \times 10^{-4})} \\[1em] \text r = \frac{4.368 \times 10^{-24}}{1.04 \times 10^{-22}} \\[1em] \text r = 4.2 \times 10^{-2}\ \text m \\[1em] \text r = 4.2\ \text{cm}\end{aligned}$$

Frequency of revolution :

$$\begin{aligned}\text n = \frac{\text{eB}}{2\text{πm}} \\[1em] \text n = \frac{(1.6 \times 10^{-19})(6.5 \times 10^{-4})}{2 \times 3.14 \times 9.1 \times 10^{-31}} \\[1em] \text n = \frac{1.04 \times 10^{-22}}{5.71 \times 10^{-30}} \\[1em] \text n = 1.8 \times 10^7\ \text{Hz} \\[1em] \text n = 18\ \text{MHz}\end{aligned}$$

No, the frequency does not depend upon the speed of the electron.

Explanation:
The expression $\text n = \frac{\text{eB}}{2\text{πm}}$ contains only e, B and m. If the speed increases, the radius increases in the same proportion, so the time taken for one revolution remains the same.

Hence, the radius of the orbit is 4.2 cm and the frequency of revolution is 18 MHz, which is independent of the speed.

Given,

• r = 0.45 m
• B = 12 T
• m = 6.7 × 10^-27 kg
• charge of α-particle, q = 2e = 3.2 × 10^-19 C

(i) From $\text r = \frac{\text{mv}}{\text{qB}}$,

$$\begin{aligned}\text v = \frac{\text{qBr}}{\text m} \\[1em] \text v = \frac{(3.2 \times 10^{-19})(12)(0.45)}{6.7 \times 10^{-27}} \\[1em] \text v = \frac{1.728 \times 10^{-18}}{6.7 \times 10^{-27}} \\[1em] \text v = 2.58 \times 10^8\ \text{ms}^{-1}\end{aligned}$$

(ii) The time-period of the particle is

$$\begin{aligned}\text T = \frac{2\text{πm}}{\text{qB}} \\[1em] \text T = \frac{2 \times 3.14 \times 6.7 \times 10^{-27}}{(3.2 \times 10^{-19})(12)} \\[1em] \text T = \frac{4.21 \times 10^{-26}}{3.84 \times 10^{-18}} \\[1em] \text T = 1.10 \times 10^{-8}\ \text s\end{aligned}$$

Hence, the speed of the particle is 2.58 × 10⁸ ms^-1 and its time-period is 1.10 × 10^-8 s.

Note: The textbook gives 1.09 × 10^-8 s; 4.21/3.84 = 1.096, which rounds to 1.10.

Given,

• v = 3.0 × 10⁷ ms^-1
• B = 0.50 N A^-1 m^-1
• m_e = 9.1 × 10^-31 kg, e = 1.6 × 10^-19 C

Radius of the circle :

$$\begin{aligned}\text r = \frac{\text{mv}}{\text{eB}} \\[1em] \text r = \frac{(9.1 \times 10^{-31})(3.0 \times 10^7)}{(1.6 \times 10^{-19})(0.50)} \\[1em] \text r = \frac{2.73 \times 10^{-23}}{8.0 \times 10^{-20}} \\[1em] \text r = 3.4 \times 10^{-4}\ \text m \\[1em] \text r = 0.34\ \text{mm}\end{aligned}$$

Kinetic energy of the electron :

$$\begin{aligned}\text K = \frac{1}{2}\text{mv}^2 \\[1em] \text K = \frac{1}{2}(9.1 \times 10^{-31})(3.0 \times 10^7)^2 \\[1em] \text K = \frac{1}{2}(9.1 \times 10^{-31})(9.0 \times 10^{14}) \\[1em] \text K = 4.1 \times 10^{-16}\ \text J\end{aligned}$$

Hence, the radius of the circle is 0.34 mm and the kinetic energy of the electron is 4.1 × 10^-16 J.

Given,

• energy of the electron, K = 40 eV = 40 × 1.6 × 10^-19 = 6.4 × 10^-18 J
• B = 9 × 10^-5 T

(i) From $\text K = \frac{1}{2}\text{mv}^2$,

$$\begin{aligned}\text v = \sqrt{\frac{2\text K}{\text m}} \\[1em] \text v = \sqrt{\frac{2 \times 6.4 \times 10^{-18}}{9.1 \times 10^{-31}}} \\[1em] \text v = \sqrt{1.407 \times 10^{13}} \\[1em] \text v = 3.75 \times 10^6\ \text{ms}^{-1}\end{aligned}$$

(ii) The radius of the circular path is

$$\begin{aligned}\text r = \frac{\text{mv}}{\text{eB}} \\[1em] \text r = \frac{(9.1 \times 10^{-31})(3.75 \times 10^6)}{(1.6 \times 10^{-19})(9 \times 10^{-5})} \\[1em] \text r = \frac{3.41 \times 10^{-24}}{1.44 \times 10^{-23}} \\[1em] \text r = 0.237\ \text m \\[1em] \text r = 23.7\ \text{cm}\end{aligned}$$

Hence, the speed of the electron is 3.75 × 10⁶ ms^-1 and the radius of the circular path is 23.7 cm.

Given,

• charge on the ion = q (mass m to be found)
• accelerating potential difference = V
• magnetic field = B, perpendicular to the plane of the figure
• the ion emerges at a distance d from the entry hole

The ion starts from rest and is accelerated through the potential difference V, so

$\frac{1}{2}\text{mv}^2 = \text{qV} \quad \Rightarrow \quad \text v = \sqrt{\frac{2\text{qV}}{\text m}} \qquad \dots(\text i)$

Inside the field region the magnetic force provides the centripetal force, so the ion describes a semi-circle of radius

$\text r = \frac{\text{mv}}{\text{qB}} \qquad \dots(\text{ii})$

The ion emerges through the hole at the bottom, at a distance d from the hole at the top, and this distance is the diameter of the semi-circle,

$\text d = 2\text r \quad \Rightarrow \quad \text r = \frac{\text d}{2}$

Substituting the value of v from (i) in (ii),

$$\begin{aligned}\frac{\text d}{2} = \frac{\text m}{\text{qB}}\sqrt{\frac{2\text{qV}}{\text m}} \\[1em] \frac{\text d}{2} = \frac{1}{\text B}\sqrt{\frac{2\text{mV}}{\text q}}\end{aligned}$$

Squaring both sides,

$$\begin{aligned}\frac{\text d^2}{4} = \frac{1}{\text B^2} \times \frac{2\text{mV}}{\text q} \\[1em] \text m = \frac{\text q \text B^2 \text d^2}{8\text V}\end{aligned}$$

Hence, the mass of the particle is $\text m = \frac{\text q \text B^2 \text d^2}{8\text V}$.

Given,

• V = 2 kV = 2000 V
• B = 0.15 T
• m_e = 9.1 × 10^-31 kg, e = 1.6 × 10^-19 C

The speed acquired by the electron is obtained from $\frac{1}{2}\text{mv}^2 = \text{eV}$,

$$\begin{aligned}\text v = \sqrt{\frac{2\text{eV}}{\text m}} \\[1em] \text v = \sqrt{\frac{2(1.6 \times 10^{-19})(2000)}{9.1 \times 10^{-31}}} \\[1em] \text v = \sqrt{7.03 \times 10^{14}} \\[1em] \text v = 2.65 \times 10^7\ \text{ms}^{-1}\end{aligned}$$

(a) When the field is transverse to the initial velocity (θ = 90°) :

The magnetic force provides the centripetal force,

$$\begin{aligned}\frac{\text{mv}^2}{\text r} = \text{qvB} \sin 90° \\[1em] \text r = \frac{\text{mv}}{\text{qB}} \\[1em] \text r = \frac{(9.1 \times 10^{-31})(2.65 \times 10^7)}{(1.6 \times 10^{-19})(0.15)} \\[1em] \text r = \frac{2.41 \times 10^{-23}}{2.4 \times 10^{-20}} \\[1em] \text r = 1.0 \times 10^{-3}\ \text m \\[1em] \text r = 1.0\ \text{mm}\end{aligned}$$

The trajectory is a circular path of radius 1.0 × 10^-3 m.

Note: The textbook's worked answer 9.95 × 10^-4 m is an arithmetic slip; the substitution gives 1.0 × 10^-3 m, as in Example 26.

(b) When the field makes an angle of 30° with the initial velocity :

In the figure, r denotes the radius r′ of the helix.

Only the component v sin θ perpendicular to the field produces the circular motion, so

$$\begin{aligned}\frac{\text m(\text v \sin 30°)^2}{\text r'} = \text q(\text v \sin 30°)\text B \\[1em] \text r' = \frac{\text{mv} \sin 30°}{\text{qB}} \\[1em] \text r' = \text r \sin 30° \\[1em] \text r' = 1.0 \times 10^{-3} \times 0.5 \\[1em] \text r' = 5.0 \times 10^{-4}\ \text m\end{aligned}$$

The trajectory is now a helical path of radius 5.0 × 10^-4 m (0.5 mm), since the component v cos 30° along the field carries the electron forward along the axis.

Note: The textbook's worked answer takes r′ = 2r; the helix radius is mv sin θ/qB = r/2, as in the chapter's theory and Example 26.

Given,

• v = 10⁷ ms^-1
• AB = 10 cm = 0.1 m, which is the diameter of the semi-circular path, so r = 0.05 m
• m_e = 9.1 × 10^-31 kg, e = 1.6 × 10^-19 C

(i) From $\text r = \frac{\text{mv}}{\text{eB}}$,

$$\begin{aligned}\text B = \frac{\text{mv}}{\text{er}} \\[1em] \text B = \frac{(9.1 \times 10^{-31})(10^7)}{(1.6 \times 10^{-19})(0.05)} \\[1em] \text B = \frac{9.1 \times 10^{-24}}{8.0 \times 10^{-21}} \\[1em] \text B = 1.12 \times 10^{-3}\ \text T\end{aligned}$$

The electron is negatively charged, so the direction of current is opposite to its motion. Applying Fleming's left-hand rule so that the force is directed towards the centre of the semi-circle, the field must be perpendicular to the plane of the paper pointing downwards.

(ii) The electron covers a semi-circle in going from A to B, so the time taken is half the period,

$$\begin{aligned}\text t = \frac{\text T}{2} \\[1em] \text t = \frac{\text{πm}}{\text{eB}} \\[1em] \text t = \frac{3.14 \times 9.1 \times 10^{-31}}{(1.6 \times 10^{-19})(1.12 \times 10^{-3})} \\[1em] \text t = \frac{2.86 \times 10^{-30}}{1.79 \times 10^{-22}} \\[1em] \text t = 1.57 \times 10^{-8}\ \text s\end{aligned}$$

Hence, the magnetic field is 1.12 × 10^-3 T, perpendicular to the plane of paper pointing downwards, and the time taken by the electron in going from A to B is 1.57 × 10^-8 s.

Given, a proton, a deuteron and an α-particle having the same kinetic energy K, entering perpendicularly a magnetic field B.

The radius of the circular path is

$$\begin{aligned}\text r = \frac{\text{mv}}{\text{qB}} \\[1em] \text r = \frac{\sqrt{2\text{mK}}}{\text{qB}}\end{aligned}$$

Since K and B are the same for all the three particles,

$\text r \propto \frac{\sqrt{\text m}}{\text q}$

Taking the mass of the proton as m and its charge as e,

• proton : m_p = m, q_p = e, so $\text r_\text p \propto \frac{\sqrt{\text m}}{\text e}$
• deuteron : m_d = 2m, q_d = e, so $\text r_\text d \propto \frac{\sqrt{2\text m}}{\text e}$
• α-particle : m_α = 4m, q_α = 2e, so $\text r_α \propto \frac{\sqrt{4\text m}}{2\text e} = \frac{\sqrt{\text m}}{\text e}$

Comparing these,

$\frac{\text r_\text d}{\text r_\text p} = \sqrt{2}, \qquad \frac{\text r_α}{\text r_\text p} = 1, \qquad \frac{\text r_\text d}{\text r_α} = \sqrt{2}$

Hence, $\text r_\text d = \sqrt{2}\ \text r_\text p$, $\text r_α = \text r_\text p$ and $\text r_\text d = \sqrt{2}\ \text r_α$.

Given,

• I = 10 A
• B = 0.5 T, in the plane of the figure and directed along AE
• BC = CD = BD = 5 cm = 0.05 m, so BCD is an equilateral triangle

(i) Force on the segment CD :

Since BCD is an equilateral triangle and BD lies along the direction of the field, the segment CD makes an angle of 60° with the field. Hence

$$\begin{aligned}\text F = \text{IB}l \sin \text θ \\[1em] \text F = (10)(0.5)(0.05)(\sin 60°) \\[1em] \text F = (10)(0.5)(0.05)(0.866) \\[1em] \text F = 0.216\ \text N\end{aligned}$$

Hence, the magnetic force on the segment CD is 0.216 N.

(ii) Net magnetic force on the part BCD :

For a current-carrying conductor placed in a uniform magnetic field, the force on a bent conductor is the same as the force on a straight conductor joining its two ends. Hence the force on BCD is equal to the force on the straight segment BD.

But BD lies along the direction of the field, so θ = 0 and

$$\begin{aligned}\text F = \text{IB}l \sin 0° \\[1em] \text F = 0\end{aligned}$$

Hence, the net magnetic force on the part BCD is zero.

Given,

• I = 8 A
• θ = 30°
• B = 0.15 T

The force on a current-carrying conductor placed in a magnetic field is

$\text F = \text{IB}l \sin \text θ$

So the force per unit length is

$$\begin{aligned}\frac{\text F}{l} = \text{IB} \sin \text θ \\[1em] \frac{\text F}{l} = (8)(0.15)(\sin 30°) \\[1em] \frac{\text F}{l} = (8)(0.15)(0.5) \\[1em] \frac{\text F}{l} = 0.6\ \text N\ \text m^{-1}\end{aligned}$$

Hence, the force per unit length on the wire is 0.6 N m^-1.

Given,

• I = 5.0 A, flowing upwards
• B = 0.020 T, horizontal and northward
• l = 0.06 m

The wire is vertical and the field is horizontal, so θ = 90° and

$$\begin{aligned}\text F = \text{IB}l \sin 90° \\[1em] \text F = (5.0)(0.020)(0.06) \\[1em] \text F = 6 \times 10^{-3}\ \text N\end{aligned}$$

Applying Fleming's left-hand rule, with the current directed vertically upwards and the field towards the north, the force on the wire is directed towards the west.

Hence, the field exerts a force of 6 × 10^-3 N on the wire, directed towards the west.

Given,

• l = 2.0 × 10^-2 m (separation of the rails)
• m = 4.0 × 10^-2 kg
• I = 3.0 A
• B = 1.2 T, directed upwards

The cylinder is horizontal and lies in the east-west direction, while the field is vertical, so θ = 90° and

$$\begin{aligned}\text F = \text{BI}l \\[1em] \text F = (1.2)(3.0)(2.0 \times 10^{-2}) \\[1em] \text F = 7.2 \times 10^{-2}\ \text N\end{aligned}$$

The positive terminal of the battery is joined to the rail on the eastern side, so the current in the cylinder flows from east to west. Applying Fleming's left-hand rule with the current towards the west and the field vertically upwards, the force on the cylinder is directed towards the north.

The acceleration of the cylinder is

$$\begin{aligned}\text a = \frac{\text F}{\text m} \\[1em] \text a = \frac{7.2 \times 10^{-2}}{4.0 \times 10^{-2}} \\[1em] \text a = 1.8\ \text{ms}^{-2}\end{aligned}$$

Hence, the magnetic force on the cylinder is 7.2 × 10^-2 N towards north, and its acceleration is 1.8 ms^-2.

Given,

• θ = 30° (angle of the inclined plane)
• B = 0.15 T, vertically directed
• mass per unit length, $\frac{\text m}{l}$ = 0.30 kg m^-1
• g = 9.8 m s^-2

The rod lies horizontally on the smooth inclined plane and the magnetic field is vertical, so the magnetic force F = IBl on the rod is horizontal. Resolving the forces along the inclined plane, the component of the magnetic force up the plane must balance the component of the weight down the plane,

$$\begin{aligned}\text{IB}l \cos \text θ = \text{mg} \sin \text θ \\[1em] \text I = \frac{\text m}{l} \times \frac{\text g \tan \text θ}{\text B}\end{aligned}$$

Substituting the values,

$$\begin{aligned}\text I = 0.30 \times \frac{(9.8)(\tan 30°)}{0.15} \\[1em] \text I = 0.30 \times \frac{(9.8)(0.577)}{0.15} \\[1em] \text I = \frac{1.697}{0.15} \\[1em] \text I = 11.3\ \text A\end{aligned}$$

Hence, the current required to keep the rod stationary is 11.3 A.

Given,

• nA = 8.0 × 10²² free electrons per metre length
• v_d = 8.0 × 10^-5 ms^-1
• B = 0.10 T, perpendicular to the conductor

(i) The current in the conductor is

$$\begin{aligned}\text I = \text{nA}\ \text v_\text d\ \text e \\[1em] \text I = (8.0 \times 10^{22})(8.0 \times 10^{-5})(1.6 \times 10^{-19}) \\[1em] \text I = 1.02\ \text A\end{aligned}$$

(ii) The force on a single electron drifting in the field is

$$\begin{aligned}\text F' = \text e\ \text v_\text d \text B \\[1em] \text F' = (1.6 \times 10^{-19})(8.0 \times 10^{-5})(0.10) \\[1em] \text F' = 1.28 \times 10^{-24}\ \text N\end{aligned}$$

(iii) The force per metre length of the conductor is

$$\begin{aligned}\frac{\text F}{l} = \text{BI} \sin 90° \\[1em] \frac{\text F}{l} = (0.10)(1.02) \\[1em] \frac{\text F}{l} = 0.1024\ \text N\ \text m^{-1}\end{aligned}$$

Hence, the current is 1.02 A, the force on an electron is 1.28 × 10^-24 N and the force per metre length of the conductor is 0.1024 N m^-1.

Given,

• l = 60 cm = 0.60 m
• m = 10 g = 0.01 kg
• B = 0.40 T
• g = 9.8 m s^-2

To neutralise the tension in the hanging wires, the magnetic force on the wire must be directed vertically upwards and must balance its weight,

$$\begin{aligned}\text{IB}l = \text{mg} \\[1em] \text I = \frac{\text{mg}}{\text B l} \\[1em] \text I = \frac{(0.01)(9.8)}{(0.40)(0.60)} \\[1em] \text I = \frac{0.098}{0.24} \\[1em] \text I = 0.41\ \text A\end{aligned}$$

The field is directed into the plane of the paper, so by Fleming's left-hand rule the force is vertically upwards only when the current flows from left to right.

Hence, a current of 0.41 A must be passed from left to right.

Given,

• I₁ = 8.0 A, I₂ = 5.0 A, in the same direction
• r = 4.0 cm = 0.04 m
• L = 10 cm = 0.10 m

The force per unit length between two parallel current-carrying conductors is

$\frac{\text F}{\text L} = \frac{\text μ_0}{2\text π}\frac{\text I_1 \text I_2}{\text r}$

So the force on a length L of the wire A is

$$\begin{aligned}\text F = \frac{\text μ_0}{2\text π}\frac{\text I_1 \text I_2}{\text r} \times \text L \\[1em] \text F = (2 \times 10^{-7}) \times \frac{(8.0)(5.0)}{0.04} \times 0.10 \\[1em] \text F = (2 \times 10^{-7})(1000)(0.10) \\[1em] \text F = 2 \times 10^{-5}\ \text N\end{aligned}$$

The currents are in the same direction, so the wires attract each other.

Hence, the force on a 10 cm section of wire A is 2 × 10^-5 N, directed towards B.

Given,

• I₁ = 100 A (lower wire), I₂ = 200 A (fine wire)
• weight per unit length of the fine wire = 0.05 N/m

The fine wire is supported by magnetic repulsion, so the force per unit length between the wires must balance its weight per unit length,

$$\begin{aligned}\frac{\text μ_0}{2\text π}\frac{\text I_1 \text I_2}{\text r} = 0.05 \\[1em] (2 \times 10^{-7}) \times \frac{(100)(200)}{\text r} = 0.05 \\[1em] \frac{4 \times 10^{-3}}{\text r} = 0.05 \\[1em] \text r = \frac{4 \times 10^{-3}}{0.05} \\[1em] \text r = 0.08\ \text m \\[1em] \text r = 8\ \text{cm}\end{aligned}$$

Hence, the fine wire should be kept 8 cm above the lower wire.

Given,

• I_A = 10 A, I_B = 20 A, I_C = 10 A, all in the same direction
• separation of A and B = 20 cm; C is placed midway between them (as shown in the figure), so r = 10 cm = 0.10 m from each
• L = 15 cm = 0.15 m

Force on C due to A :

$$\begin{aligned}\text F_1 = \frac{\text μ_0}{2\text π}\frac{\text I_\text A \text I_\text C}{\text r}\text L \\[1em] \text F_1 = (2 \times 10^{-7}) \times \frac{(10)(10)}{0.10} \times 0.15 \\[1em] \text F_1 = 3 \times 10^{-5}\ \text N,\ \text{directed towards A}\end{aligned}$$

Force on C due to B :

$$\begin{aligned}\text F_2 = \frac{\text μ_0}{2\text π}\frac{\text I_\text B \text I_\text C}{\text r}\text L \\[1em] \text F_2 = (2 \times 10^{-7}) \times \frac{(20)(10)}{0.10} \times 0.15 \\[1em] \text F_2 = 6 \times 10^{-5}\ \text N,\ \text{directed towards B}\end{aligned}$$

The currents in all the three wires are in the same direction, so C is attracted by both A and B. The two forces are oppositely directed, so the net force is

$$\begin{aligned}\text F = \text F_2 - \text F_1 \\[1em] \text F = (6 - 3) \times 10^{-5} \\[1em] \text F = 3.0 \times 10^{-5}\ \text N\end{aligned}$$

Hence, a force of 3.0 × 10^-5 N acts on C, directed towards B.

Given,

• applied frequency, ν₀ = 8 megacycles/second = 8 × 10⁶ Hz
• m_p = 1.67 × 10^-27 kg
• q = 1.6 × 10^-19 C

For resonance in a cyclotron, the frequency of the applied potential difference must equal the frequency of revolution of the particle,

$\text ν_0 = \frac{\text{qB}}{2\text{πm}} \quad \Rightarrow \quad \text B = \frac{2\text{πm}\text ν_0}{\text q}$

Substituting the values,

$$\begin{aligned}\text B = \frac{2 \times 3.14 \times (1.67 \times 10^{-27})(8 \times 10^6)}{1.6 \times 10^{-19}} \\[1em] \text B = \frac{8.39 \times 10^{-20}}{1.6 \times 10^{-19}} \\[1em] \text B = 0.524\ \text T\end{aligned}$$

Hence, the cyclotron dees should be placed in a magnetic field of 0.524 T.

Given,

• R = 0.4 m (radius of the dees)
• ν₀ = 12 megacycles/second = 12 × 10⁶ Hz
• V = 5000 volt (maximum value)
• m_p = 1.67 × 10^-27 kg, q = 1.6 × 10^-19 C

Magnetic field : For resonance,

$$\begin{aligned}\text ν_0 = \frac{\text{qB}}{2\text{πm}} \quad \Rightarrow \quad \text B = \frac{2\text{πm}\text ν_0}{\text q} \\[1em] \text B = \frac{2 \times 3.14 \times (1.67 \times 10^{-27})(12 \times 10^6)}{1.6 \times 10^{-19}} \\[1em] \text B = 0.786\ \text T\end{aligned}$$

Velocity acquired : At the outer edge of a dee, r = R, so

$$\begin{aligned}\text v_\text{max} = \frac{\text{qBR}}{\text m} \\[1em] \text v_\text{max} = \frac{(1.6 \times 10^{-19})(0.786)(0.4)}{1.67 \times 10^{-27}} \\[1em] \text v_\text{max} = \frac{5.03 \times 10^{-20}}{1.67 \times 10^{-27}} \\[1em] \text v_\text{max} = 3 \times 10^7\ \text{ms}^{-1}\end{aligned}$$

Energy acquired :

$$\begin{aligned}\text K = \frac{1}{2}\text m \text v_\text{max}^2 \\[1em] \text K = \frac{1}{2}(1.67 \times 10^{-27})(3 \times 10^7)^2 \\[1em] \text K = 7.56 \times 10^{-13}\ \text J \\[1em] \text K = \frac{7.56 \times 10^{-13}}{1.6 \times 10^{-19}}\ \text{eV} \\[1em] \text K = 4.73\ \text{MeV}\end{aligned}$$

Hence, the required magnetic field is 0.786 T, the energy acquired by the protons is 4.73 MeV and their velocity is 3 × 10⁷ ms^-1.

Given,

• v = 2.0 × 10⁶ ms^-1, vertically upwards
• B_H = 0.32 × 10^-4 T, horizontal and towards the north
• e = 1.6 × 10^-19 C

The electron moves vertically while the horizontal component of the earth's field is towards the north, so θ = 90° and

$$\begin{aligned}\text F = \text{B}_\text H \text{ev} \\[1em] \text F = (0.32 \times 10^{-4})(1.6 \times 10^{-19})(2.0 \times 10^6) \\[1em] \text F = 1.024 \times 10^{-17}\ \text N\end{aligned}$$

The electron is negatively charged, so the direction of current is vertically downwards. Applying Fleming's left-hand rule with the current directed downwards and the field towards the north, the force on the electron is directed towards the east.

Hence, the force on the electron is 1.024 × 10^-17 N, directed towards the east.

The magnetic field at P, produced by the current I₂ in the conductor 2, is

$\text B_{12} = \frac{\text µ_0}{2\text π}\frac{\text I_2}{\text d}$

and the magnetic field at Q, produced by the current I₁ in the conductor 1, is

$\text B_{21} = \frac{\text µ_0}{2\text π}\frac{\text I_1}{\text d}$

Directions : By the right-hand thumb rule, both the conductors carrying currents in the same direction, $\vec{\text B}_{12} \space$at P is perpendicular to the plane of the paper directed outwards, while $\vec{\text B}_{21} \space$at Q is perpendicular to the plane of the paper directed inwards.

Definition of the SI unit of current : Each conductor lies in the field of the other and therefore experiences a force. The force per unit length is

$\frac{\text F}{\text L} = \frac{\text µ_0}{2\text π}\frac{\text I_1 \text I_2}{\text d}$

Putting I₁ = I₂ = 1 A and d = 1 m,

$$\begin{aligned}\frac{\text F}{\text L} = \frac{4\text π \times 10^{-7}}{2\text π} \\[1em] \frac{\text F}{\text L} = 2 \times 10^{-7}\ \text{N m}^{-1}\end{aligned}$$

Hence, one ampere is that steady current which, when maintained in each of two infinitely long, straight and parallel conductors of negligible cross-section placed in vacuum at a distance of 1 metre from each other, produces a force of 2 × 10^-7 newton per metre length on each conductor.

Let a particle of mass m and charge q enter a uniform magnetic field $\vec{\text B} \space$with a velocity $\vec{\text v} \space$directed perpendicular to the field.

Since $\vec{\text v} \space$is perpendicular to $\vec{\text B} \space$, the particle is acted upon by a force of magnitude qvB which lies in the plane of the path and is always perpendicular to the velocity. Such a force does not change the magnitude of the velocity, it changes only its direction. Hence the particle moves along a circle of radius r, the magnetic force acting as the centripetal force,

$\text{qvB} = \frac{\text{mv}^2}{\text r}$

Solving for r,

$\text r = \frac{\text{mv}}{\text{qB}}$

But mv is the momentum p of the particle, so

$\text r = \frac{\text p}{\text{qB}}$

For a given particle in a given field, q and B are constants, so

$\text r \propto \text p$

Hence, the radius of the path followed by a charged particle entering perpendicularly in a magnetic field is directly proportional to its momentum.

Given,

• l = 10 cm = 0.1 m
• I = 10 A
• θ = 53°, with sin 53° = 4/5
• B = 0.1 tesla

The force experienced by a current-carrying conductor in a magnetic field is

$\text F = \text{IB}l \sin \text θ$

Substituting the values,

$$\begin{aligned}\text F = (10)(0.1)(0.1)\left(\frac{4}{5}\right) \\[1em] \text F = 0.1 \times 0.8 \\[1em] \text F = 0.08\ \text N\end{aligned}$$

Hence, the force experienced by the wire is 0.08 N.

Given,

• I = 1.5 A in a long solenoid of 500 turns
• v = 2.0 × 10⁶ ms^-1, along the axis of the solenoid

Inside a long solenoid the magnetic field is uniform and is directed along the axis of the solenoid. The electron is moving along this very axis, so the angle between $\vec{\text v} \space$and $\vec{\text B} \space$is zero.

$$\begin{aligned}\text F = \text{Bev} \sin \text θ \\[1em] \text F = \text{Bev} \sin 0° \\[1em] \text F = 0\end{aligned}$$

Hence, the magnetic force on the electron is zero.

Given,

• I = 5 A in the wire PQ
• v = 10⁶ m/s, in the same direction as the current
• r = 10 cm = 0.1 m

The magnetic field produced by the wire at the position of the electron is

$$\begin{aligned}\text B = \frac{\text µ_0}{4\text π}\frac{2\text I}{\text r} \\[1em] \text B = (10^{-7}) \times \frac{2 \times 5}{0.1} \\[1em] \text B = 1 \times 10^{-5}\ \text T\end{aligned}$$

This field is perpendicular to the plane of the paper, so the electron moves at right angles to it and

$$\begin{aligned}\text F = \text{evB} \\[1em] \text F = (1.6 \times 10^{-19})(10^6)(1 \times 10^{-5}) \\[1em] \text F = 1.6 \times 10^{-18}\ \text N\end{aligned}$$

The electron is negatively charged, so the direction of current is opposite to its motion. Applying Fleming's left-hand rule, the force on the electron is directed towards the right, that is, away from the wire.

Hence, the force experienced by the electron is 1.6 × 10^-18 N, towards right (away from the wire).

Ampere's circuital law : The line integral of the magnetic field (magnetic flux density) $\vec{\text B} \space$around any closed path in free space is equal to µ₀ times the net current I threading through the area enclosed by the path,

$\oint \vec{\text B} \cdot \vec{\text{dl}} = \text µ_0 \text I$

where µ₀ is the permeability of free space and I is the steady current that does not change with time. The law holds for any assembly of currents and for any closed path.

(i) In vector form, Biot-Savart's law states that the magnetic field produced at a point by a current-element $\text I \vec{\text{dl}} \space$is

$\text d\vec{\text B} = \frac{\text µ_0}{4\text π}\frac{\text I(\vec{\text{dl}} \times \vec{\text r})}{\text r^3}$

where $\vec{\text r} \space$is the vector drawn from the current-element to the point.

The magnetic field at the centre of a circular coil of radius R having N turns and carrying a current I is

$\text B = \frac{\text µ_0 \text N \text I}{2\text R}$

(ii) Given,

• R = 0.8 m
• time for one round, T = 2 s
• charge of the helium nucleus, q = 2e = 3.2 × 10^-19 C

The revolving nucleus is equivalent to a current loop, and the equivalent current is

$$\begin{aligned}\text I = \frac{\text q}{\text T} \\[1em] \text I = \frac{3.2 \times 10^{-19}}{2} \\[1em] \text I = 1.6 \times 10^{-19}\ \text A\end{aligned}$$

The magnetic field at the centre of the circle is

$$\begin{aligned}\text B = \frac{\text µ_0 \text I}{2\text R} \\[1em] \text B = \frac{(4\text π \times 10^{-7})(1.6 \times 10^{-19})}{2 \times 0.8} \\[1em] \text B = \frac{2.01 \times 10^{-25}}{1.6} \\[1em] \text B = 12.56 \times 10^{-26}\ \text T\end{aligned}$$

Hence, the magnetic field at the centre of the circle is 12.56 × 10^-26 T.

$\vec{\text F} = \text q(\vec{\text v} \times \vec{\text B})$

Explanation:
The magnetic field B at a point is defined through the force experienced by a charge moving at that point, where q is the charge and $\vec{\text v} \space$its velocity. The magnitude of this force is

$$\begin{aligned}\text F = \text{Bqv} \sin \text θ \\[1em] \text B = \frac{\text F}{\text{qv} \sin \text θ}\end{aligned}$$

θ being the angle between $\vec{\text v} \space$and $\vec{\text B} \space$.

Given,

• E = 3.4 × 10⁴ Vm^-1
• B = 2 × 10^-3 Wbm^-2
• m_e = 9.1 × 10^-31 kg, e = 1.6 × 10^-19 C

Speed of the electrons : The path of the beam remains undeviated, so the electric force balances the magnetic force,

$$\begin{aligned}\text{eE} = \text{evB} \\[1em] \text v = \frac{\text E}{\text B} \\[1em] \text v = \frac{3.4 \times 10^4}{2 \times 10^{-3}} \\[1em] \text v = 1.7 \times 10^7\ \text{m/s}\end{aligned}$$

Radius of the circular path : When the electric field is removed, only the magnetic force acts and the electrons describe a circle of radius

$$\begin{aligned}\text r = \frac{\text{mv}}{\text{eB}} \\[1em] \text r = \frac{(9.1 \times 10^{-31})(1.7 \times 10^7)}{(1.6 \times 10^{-19})(2 \times 10^{-3})} \\[1em] \text r = \frac{1.547 \times 10^{-23}}{3.2 \times 10^{-22}} \\[1em] \text r = 4.83 \times 10^{-2}\ \text m\end{aligned}$$

Hence, the speed of the electrons is 1.7 × 10⁷ m/s and the radius of the circular path is 4.83 × 10^-2 m.

$\vec{\text F} = \text q(\vec{\text v} \times \vec{\text B})$

Explanation:
This is the Lorentz force acting on a charged particle having charge q moving with velocity $\vec{\text v} \space$in a magnetic field $\vec{\text B} \space$. Its magnitude is F = Bqv sin θ, where θ is the angle between $\vec{\text v} \space$and $\vec{\text B} \space$, and its direction is perpendicular to the plane containing $\vec{\text v} \space$and $\vec{\text B} \space$.

When an electric field $\vec{\text E} \space$is also present, the complete Lorentz force is

$\vec{\text F} = \text q(\vec{\text E} + \vec{\text v} \times \vec{\text B})$

Biot-Savart's law : The magnetic field dB developed at a point due to a small current-element I dl is directly proportional to the current I, to the length dl of the element and to sin θ, where θ is the angle between the element and the line joining it to the point, and inversely proportional to the square of the distance r between them,

$\text{dB} = \frac{\text µ_0}{4\text π}\frac{\text I\ \text{dl} \sin \text θ}{\text r^2}$

Magnetic field at the centre of a circular coil :

Let a circular coil of radius r carry a current I, and let O be its centre. Consider a small current-element dl of the coil. Every element of a circle is perpendicular to the radius drawn to the centre, so θ = 90° and the field at O due to this element is

$$\begin{aligned}\text{dB} = \frac{\text µ_0}{4\text π}\frac{\text I\ \text{dl} \sin 90°}{\text r^2} \\[1em] \text{dB} = \frac{\text µ_0}{4\text π}\frac{\text I\ \text{dl}}{\text r^2}\end{aligned}$$

The direction of the field is the same for all the elements of the coil, so the fields simply add up. Hence the field at O due to the entire coil is

$\text B = \frac{\text µ_0}{4\text π}\frac{\text I}{\text r^2}\sum \text{dl}$

But $\sum \text{dl} = 2\text{πr}$, the circumference of the coil, so

$$\begin{aligned}\text B = \frac{\text µ_0}{4\text π}\frac{\text I}{\text r^2}(2\text{πr}) \\[1em] \text B = \frac{\text µ_0 \text I}{2\text r}\end{aligned}$$

If the coil has N turns, each turn contributes equally, so

$\text B = \frac{\text µ_0 \text N \text I}{2\text r}$

The conductor should be placed perpendicular to the field.

Explanation:
The force acting on a current-carrying conductor placed in a magnetic field is

$\text F = \text{IB}l \sin \text θ$

which is maximum when sin θ = 1, that is when θ = 90°. Then F_max = IBl.

its speed remains constant

Reason — The magnetic force on the particle is always perpendicular to its direction of motion, so it does no work on the particle. Hence the speed, and with it the kinetic energy, remains constant.

Let the long conductor carry a steady current I, and let P be a point at a perpendicular distance a from it.

Draw a circle of radius a through P, with its centre on the conductor and its plane perpendicular to the conductor. By symmetry, the magnitude of $\vec{\text B} \space$is the same at all points of this circle and its direction is tangential to the circle at each point. Hence $\vec{\text B} \space$and $\vec{\text{dl}} \space$are along the same direction at every element, and

$$\begin{aligned}\oint \vec{\text B} \cdot \vec{\text{dl}} = \oint \text B\ \text{dl} \cos 0° \\[1em] \oint \vec{\text B} \cdot \vec{\text{dl}} = \text B \oint \text{dl} \\[1em] \oint \vec{\text B} \cdot \vec{\text{dl}} = \text B(2\text{πa})\end{aligned}$$

The current enclosed by the circle is I, so applying Ampere's circuital law,

$$\begin{aligned}\text B(2\text{πa}) = \text µ_0 \text I \\[1em] \text B = \frac{\text µ_0 \text I}{2\text{πa}} \\[1em] \text B = \left(\frac{\text µ_0}{4\text π}\right)\frac{2\text I}{\text a}\end{aligned}$$

Hence, the magnetic flux density at P is $\text B = \left(\frac{\text µ_0}{4\text π}\right)\frac{2\text I}{\text a}$.

$\frac{\text F}{\text L} = \frac{\text µ_0}{2\text π}\frac{\text I_1 \text I_2}{\text r}$

Explanation:
Each conductor lies in the magnetic field produced by the current in the other, and so experiences a force.

If the conductors X and Y carry currents I₁ and I₂ and are kept parallel at a distance r apart in vacuum, the field produced by Y at the position of X is

$\text B = \frac{\text µ_0}{2\text π}\frac{\text I_2}{\text r}$

The conductor X lies at right angles to this field, so the force on a length L of it is F = I₁BL. Dividing by L gives the force per unit length stated above.

This force is one of attraction if the currents are in the same direction, and one of repulsion if they are in opposite directions.

Given,

• v = 2 × 10⁶ ms^-1
• E = 5 × 10⁴ Vm^-1

(i) The electrons pass undeviated, so the electric force balances the magnetic force,

$$\begin{aligned}\text{eE} = \text{evB} \\[1em] \text B = \frac{\text E}{\text v} \\[1em] \text B = \frac{5 \times 10^4}{2 \times 10^6} \\[1em] \text B = 2.5 \times 10^{-2}\ \text T\end{aligned}$$

(ii) The three vectors $\vec{\text v} \space$, $\vec{\text E} \space$and $\vec{\text B} \space$are mutually perpendicular. With $\vec{\text v} \space$towards the right and $\vec{\text E} \space$vertically downwards in the plane of the paper, $\vec{\text B} \space$must be perpendicular to the plane of the paper. For the two forces on the electron to be opposite, the field must be directed into the plane of the paper (inwards).

Hence, the magnetic flux density is 2.5 × 10^-2 T, directed perpendicular to the plane of paper and inwards.

(i) The current I in the conductor XY flows upwards, and the point P lies to the right of XY. Applying the right-hand thumb rule, the magnetic flux density B at P is perpendicular to the plane of the paper, directed into the page.

(ii) The short conductor MN at P carries a current I in the same direction as the current in XY, so the two currents are parallel. Parallel currents attract each other, so the force experienced by MN is directed towards the conductor XY.

The same result follows from Fleming's left-hand rule, with the field at P into the page and the current in MN directed upwards.

(i) The field produced by the conductor X at the position of the conductor Y is

$\text B = \left(\frac{\text µ_0}{4\text π}\right)\frac{2\text I_1}{\text a}$

The conductor Y lies at right angles to this field, so the force on a length L of Y is F = I₂BL, and the force per unit length is

$\frac{\text F}{\text L} = \left(\frac{\text µ_0}{4\text π}\right)\frac{2\text I_1 \text I_2}{\text a}$

(ii) Definition of one ampere : One ampere is that steady current which, when maintained in each of two infinitely long, straight and parallel conductors of negligible cross-section placed in vacuum at a distance of 1 metre from each other, produces a force of 2 × 10^-7 newton per metre length on each conductor.

$\delta\vec{\text B} = \frac{\text µ_0}{4\text π}\frac{\text I(\vec{\text{dl}} \times \vec{\text r})}{\text r^3}$

Reason — In vector form Biot-Savart's law is written with the current-element as the vector $\text I \vec{\text{dl}}$, and the field is perpendicular to the plane containing $\vec{\text{dl}} \space$and $\vec{\text r} \space$. Writing $\vec{\text r} = \text r\hat{\text r}$, the scalar form

$\delta \text B = \frac{\text µ_0}{4\text π}\frac{\text I\ \text{dl} \sin \text θ}{\text r^2}$

becomes $\delta\vec{\text B} = \frac{\text µ_0}{4\text π}\frac{\text I(\vec{\text{dl}} \times \vec{\text r})}{\text r^3}$, with r³ in the denominator because the numerator carries one extra factor of r.

neutrons

Reason — Both a cyclotron and a van de Graaff generator accelerate a particle by means of an electric field, and the force exerted by an electric field on a particle is F = qE. A neutron carries no charge, so no electric force acts on it and it cannot be accelerated by either machine.

10 A

Reason — Given, l = 2 m, weight W = 4 × 10^-3 N and B = 2 × 10^-4 T.

The wire remains at rest, so the magnetic force balances its weight,

$$\begin{aligned}\text{BI}l = \text W \\[1em] \text I = \frac{\text W}{\text B l} \\[1em] \text I = \frac{4 \times 10^{-3}}{(2 \times 10^{-4})(2)} \\[1em] \text I = \frac{4 \times 10^{-3}}{4 \times 10^{-4}} \\[1em] \text I = 10\ \text A\end{aligned}$$

A moving charged particle experiences no force while moving through a uniform magnetic field when it moves parallel or anti-parallel to the direction of the field.

Explanation:
The magnetic force on a charged particle is

$\text F = \text{Bqv} \sin \text θ$

where θ is the angle between $\vec{\text v} \space$and $\vec{\text B} \space$. This force is zero when sin θ = 0, that is when θ = 0° or 180°.

double

Reason — The magnetic field at the centre of a circular coil is

$\text B = \frac{\text µ_0 \text N \text I}{2\text R}$

When the current, the radius and the number of turns are all doubled, that is I → 2I, R → 2R and N → 2N,

$$\begin{aligned}\text B' = \frac{\text µ_0 (2\text N)(2\text I)}{2(2\text R)} \\[1em] \text B' = 2 \times \frac{\text µ_0 \text N \text I}{2\text R} \\[1em] \text B' = 2\text B\end{aligned}$$

The force acting on q will be maximum when it moves perpendicular to the magnetic field B, that is, when θ = 90°.

Explanation:
The magnetic force on the charge is

$\text F = \text{Bqv} \sin \text θ$

which is maximum when sin θ = 1, that is θ = 90°, giving F_max = Bqv. As the particle travels along the positive x-axis, the force is maximum when the field B is directed at right angles to the x-axis.

Given,

• I₁ = 50 A in the wire P
• I₂ = 25 A in the wire Q
• weight per unit length of Q = 0.025 Nm^-1

The wire Q remains at rest above P, so the magnetic force per unit length on it must be directed upwards and must balance its weight per unit length. From the figure the two currents are in opposite directions, so the wires repel each other and Q is supported.

$$\begin{aligned}\frac{\text µ_0}{2\text π}\frac{\text I_1 \text I_2}{\text r} = 0.025 \\[1em] (2 \times 10^{-7}) \times \frac{(50)(25)}{\text r} = 0.025 \\[1em] \frac{2.5 \times 10^{-4}}{\text r} = 0.025 \\[1em] \text r = \frac{2.5 \times 10^{-4}}{0.025} \\[1em] \text r = 0.01\ \text m = 1\ \text{cm}\end{aligned}$$

Hence, the wire Q should be kept at a distance of 1 cm from the wire P.

Given,

• I₁ = 75 A in the wire A
• I₂ = 20 A in the wire B
• r = 1 cm = 0.01 m

The force per unit length acting on the wire B due to the current in the wire A is

$\frac{\text F}{\text L} = \frac{\text µ_0}{2\text π}\frac{\text I_1 \text I_2}{\text r}$

Substituting the values,

$$\begin{aligned}\frac{\text F}{\text L} = (2 \times 10^{-7}) \times \frac{(75)(20)}{0.01} \\[1em] \frac{\text F}{\text L} = (2 \times 10^{-7})(1.5 \times 10^5) \\[1em] \frac{\text F}{\text L} = 3 \times 10^{-2}\ \text{N/m}\end{aligned}$$

The currents in the two wires are in opposite directions, so the force is one of repulsion, directed away from the wire A.

Hence, the force per unit length acting on the wire B is 3 × 10^-2 N/m.

(a)

For an infinitely long straight conductor carrying a current I, the magnetic field at a point P at a perpendicular distance r from it is

$\text B = \frac{\text µ_0}{2\text π}\frac{\text I}{\text r}$

The current flows upwards and P lies to the right of the conductor, so by the right-hand thumb rule the field at P is perpendicular to the plane of the paper, directed into the page.

(b)

For a circular loop of radius R carrying a current I, the magnetic field at the centre is

$\text B = \frac{\text µ_0 \text I}{2\text R}$

The current flows in the anticlockwise sense, so by the right-hand palm rule no. 1 the field at the centre is perpendicular to the plane of the loop, directed out of the page (towards the observer).

$\text d\vec{\text B} = \frac{\text µ_0}{4\text π}\frac{\text I(\vec{\text{dl}} \times \vec{\text r})}{\text r^3}$

Explanation:
Here $\text d\vec{\text B} \space$is the magnetic field produced at a point by a current-element $\text I \vec{\text{dl}}$, $\vec{\text r} \space$is the vector drawn from the current-element to the point at which the field is required, and µ₀ is the permeability of free space. The direction of $\text d\vec{\text B} \space$is perpendicular to the plane containing $\vec{\text{dl}} \space$and $\vec{\text r} \space$.

If two long conductors carrying currents I₁ and I₂ are kept parallel to each other at a distance r apart in vacuum, the force per unit length between them is

$\frac{\text F}{\text L} = \frac{\text µ_0}{2\text π}\frac{\text I_1 \text I_2}{\text r}$

This force is attractive if the currents are in the same direction and repulsive if they are in opposite directions.

Definition of ampere : Putting I₁ = I₂ = 1 A and r = 1 m,

$$\begin{aligned}\frac{\text F}{\text L} = \frac{4\text π \times 10^{-7}}{2\text π} \times \frac{1 \times 1}{1} \\[1em] \frac{\text F}{\text L} = 2 \times 10^{-7}\ \text{N m}^{-1}\end{aligned}$$

Hence, one ampere is that steady current which, when maintained in each of two infinitely long, straight and parallel conductors of negligible cross-section placed in vacuum at a distance of 1 metre from each other, produces a force of 2 × 10^-7 newton per metre length on each conductor.

B = µ₀nI

Reason — Applying Ampere's circuital law to a rectangular path partly inside and partly outside a long solenoid gives Bx = µ₀(nx)I, so

$\text B = \text µ_0 \text n \text I$

where n is the number of turns per unit length.

$\vec{\text F} = \text q(\vec{\text E} + \vec{\text v} \times \vec{\text B})$

Explanation:
The Lorentz force is the total force acting on a charged particle moving in a region where both an electric field $\vec{\text E} \space$and a magnetic field $\vec{\text B} \space$are present. Here q is the charge on the particle and $\vec{\text v} \space$its velocity. The first term is the electric force, which acts whether the charge is at rest or in motion, and the second term is the magnetic force, which acts only on a moving charge.

0°

Reason — The force on the proton is

$\text F = \text{Bqv} \sin \text θ$

which is minimum when sin θ is minimum. Among the given angles sin θ is least at θ = 0°, where sin 0° = 0 and the force becomes zero. At θ = 90° the force is maximum.

move towards the wire P

Reason — The two wires carry currents in the same direction, so they attract each other. The force per unit length between them is

$\frac{\text F}{\text L} = \frac{\text µ_0}{2\text π}\frac{\text I_1 \text I_2}{\text r}$

directed towards the other wire. As the wire P is kept fixed, the wire Q moves towards P.

B = µ₀NI/2r

Reason — Applying Biot-Savart's law to every element of the coil and adding the fields, which are all in the same direction at the centre,

$$\begin{aligned}\text B = \frac{\text µ_0}{4\text π}\frac{\text I}{\text r^2}(2\text{πr}) \\[1em] \text B = \frac{\text µ_0 \text I}{2\text r}\end{aligned}$$

for a single turn. For N turns each turn contributes equally, so $\text B = \frac{\text µ_0 \text{NI}}{2\text r}$.

Given,

• I = 5 A in the wire AB
• v = 2 × 10⁶ m/s, parallel to the wire but opposite to the current
• r = 0.2 m

The magnetic field produced by the conductor at the position of the proton is

$$\begin{aligned}\text B = \frac{\text µ_0}{4\text π}\frac{2\text I}{\text r} \\[1em] \text B = (10^{-7}) \times \frac{2 \times 5}{0.2} \\[1em] \text B = 5 \times 10^{-6}\ \text T\end{aligned}$$

This field is perpendicular to the plane of the paper, so the proton moves at right angles to it and

$$\begin{aligned}\text F = \text{qvB} \\[1em] \text F = (1.6 \times 10^{-19})(2 \times 10^6)(5 \times 10^{-6}) \\[1em] \text F = 1.6 \times 10^{-18}\ \text N\end{aligned}$$

Applying Fleming's left-hand rule, this force is directed towards the right hand side, that is, away from the wire.

Hence, the force exerted on the proton is 1.6 × 10^-18 N, towards right hand side (away from the wire).

Let an infinitely long straight conductor carry a steady current I, and let P be a point at a perpendicular distance r from it.

Draw a circle of radius r through P, with its centre on the conductor and its plane perpendicular to the conductor. By symmetry, the magnitude of $\vec{\text B} \space$is the same at all points of this circle and its direction is tangential to the circle at each point. Hence $\vec{\text B} \space$and $\vec{\text{dl}} \space$are along the same direction at every element of the circle, and

$$\begin{aligned}\oint \vec{\text B} \cdot \vec{\text{dl}} = \oint \text B\ \text{dl} \cos 0° \\[1em] \oint \vec{\text B} \cdot \vec{\text{dl}} = \text B \oint \text{dl}\end{aligned}$$

But $\oint \text{dl} = 2\text{πr}$, the circumference of the circle, and the current enclosed by the circle is I. Applying Ampere's circuital law,

$$\begin{aligned}\text B(2\text{πr}) = \text µ_0 \text I \\[1em] \text B = \frac{\text µ_0}{2\text π}\frac{\text I}{\text r}\end{aligned}$$

Hence, the magnetic flux density at a point near an infinitely long straight conductor is $\text B = \frac{\text µ_0}{2\text π}\frac{\text I}{\text r}$.

$\delta\vec{\text B} = \text{KI}{\vec{\text{dl}} \times \vec{\text r}}/\text r^3$

Reason — Biot-Savart's law in scalar form is

$\delta \text B = \text{KI}\frac{\text{dl} \sin \text θ}{\text r^2}, \qquad \text K = \frac{\text µ_0}{4\text π}$

Since $|\vec{\text{dl}} \times \vec{\text r}| = \text{dl}\ \text r \sin \text θ$, writing the numerator as a cross product introduces one extra factor of r, which must be compensated by r³ in the denominator.

Given,

• l = 20 cm = 0.2 m
• m = 1.2 g = 1.2 × 10^-3 kg
• B = 0.3 T
• g = 9.8 m s^-2

The wire does not fall, so the magnetic force on it balances its weight,

$$\begin{aligned}\text{BI}l = \text{mg} \\[1em] \text I = \frac{\text{mg}}{\text B l} \\[1em] \text I = \frac{(1.2 \times 10^{-3})(9.8)}{(0.3)(0.2)} \\[1em] \text I = \frac{11.76 \times 10^{-3}}{0.06} \\[1em] \text I = 0.196\ \text A\end{aligned}$$

Hence, a current of 0.196 A (that is, 196 × 10^-3 A) should be passed through the wire.

Note: The printed answer key gives 19.6 × 10^-3 A. Substituting the given data in BIl = mg yields 196 × 10^-3 A, so the printed value appears to be smaller by a factor of ten.

$\vec{\text F} = \text q(\vec{\text v} \times \vec{\text B})$

Reason — The Lorentz force on a charge q moving with velocity $\vec{\text v} \space$in a magnetic field $\vec{\text B} \space$is the vector product of $\vec{\text v} \space$and $\vec{\text B} \space$multiplied by q, taken in that order.

length = x, area of cross section = 2y

Reason — The resistance of a wire is

$\text R = \text ρ\frac{l}{\text A}$

so R is least when the length is least and the area of cross-section is greatest. Among the four wires, the one with length x and area 2y gives

$\text R = \frac{\text ρ \text x}{2\text y}$

which is the smallest of the four values $\frac{\text ρ \text x}{\text y}$, $\frac{2\text ρ \text x}{\text y}$, $\frac{\text ρ \text x}{2\text y}$ and $\frac{\text ρ \text x}{\text y}$.

Note: The printed answer key gives option 1 (length = x, area = y), but that wire has resistance $\frac{\text ρ \text x}{\text y}$, which is twice the resistance of the wire of option 3. Hence option 3 is the correct choice.

(i) The current I' in the wire YY' flows upwards and the point P lies to the right of the wire. By the right-hand thumb rule, the magnetic flux density B at P is perpendicular to the plane of the paper and directed downwards (into the page).

(ii) For an infinitely long straight conductor, the magnitude of the magnetic flux density at a perpendicular distance r is

$\text B = \frac{\text µ_0}{2\text π}\frac{\text I'}{\text r}$

(iii) The wire MN of length l carrying a current I is placed at P, parallel to YY', and lies at right angles to the field B. Hence the force on it is

$$\begin{aligned}\text F = \text{BI}l \\[1em] \text F = \frac{\text µ_0}{2\text π}\frac{\text I'\ \text I\ l}{\text r}\end{aligned}$$

This force is one of attraction if the two currents are in the same direction, and one of repulsion if they are in opposite directions.

Let two long straight wires be kept parallel to each other in vacuum at a distance r apart, carrying currents I and I' respectively.

The magnitude of the magnetic field produced by the current I at the position of the second wire is

$\text B = \frac{\text µ_0}{2\text π}\frac{\text I}{\text r}$

and by the right-hand palm rule no. 1 this field is perpendicular to the plane containing the two wires.

The second wire, carrying the current I', is thus situated in this field at right angles to it, so the force on a length L of it is

$$\begin{aligned}\text F = \text I' \text B \text L \sin 90° \\[1em] \text F = \text I'\left(\frac{\text µ_0}{2\text π}\frac{\text I}{\text r}\right)\text L\end{aligned}$$

Therefore the force per unit length is

$$\begin{aligned}\text F' = \frac{\text F}{\text L} \\[1em] \text F' = \frac{\text µ_0}{2\text π}\frac{\text I\ \text I'}{\text r}\end{aligned}$$

Writing $\frac{\text µ_0}{2\text π} = \left(\frac{\text µ_0}{4\text π}\right)\times 2$,

$\text F' = \left(\frac{\text µ_0}{4\text π}\right)\frac{2\text{II}'}{\text r}$

Hence, the force per unit length between the two wires is $\text F' = \left(\frac{\text µ_0}{4\text π}\right)\frac{2\text{II}'}{\text r}$, where the terms have their usual meaning.

1 MeV

Reason — Given, the radii of the circular orbits are equal and K_p = 1 MeV.

For a particle entering perpendicular to the field,

$\text r = \frac{\sqrt{2\text{mK}}}{\text{qB}}$

As the radii are equal,

$\frac{\sqrt{2\text m_\text p \text K_\text p}}{\text e} = \frac{\sqrt{2\text m_α \text K_α}}{2\text e}$

Putting m_α = 4m_p,

$$\begin{aligned}\sqrt{\text m_\text p \text K_\text p} = \frac{\sqrt{4\text m_\text p \text K_α}}{2} \\[1em] \sqrt{\text m_\text p \text K_\text p} = \sqrt{\text m_\text p \text K_α} \\[1em] \text K_α = \text K_\text p = 1\ \text{MeV}\end{aligned}$$

$\frac{\text µ_0 \text{ne}}{2\text r}$

Reason — Given, the electron makes n revolutions per second in a circle of radius r.

The revolving electron is equivalent to a current loop, and the equivalent current is

$$\begin{aligned}\text I = \frac{\text e}{\text T} \\[1em] \text I = \text{ne}\end{aligned}$$

The magnetic field at the centre of a circular loop is

$$\begin{aligned}\text B = \frac{\text µ_0 \text I}{2\text r} \\[1em] \text B = \frac{\text µ_0 \text{ne}}{2\text r}\end{aligned}$$

$\vec{\text B} = -\frac{\text µ_0}{4\text π}\frac{\text I}{\text R}(\text π\hat{\text i} + 2\hat{\text k})$

Reason — Given, two very long linear parts parallel to the X-axis and a semi-circular portion of radius R lying in the Y-Z plane.

Field due to the linear parts : Each linear part is a semi-infinite conductor whose perpendicular distance from O is R, so each gives

$$\begin{aligned}\text B_1 = \frac{\text µ_0}{4\text π}\frac{\text I}{\text R}(\sin 0° + \sin 90°) \\[1em] \text B_1 = \frac{\text µ_0}{4\text π}\frac{\text I}{\text R}\end{aligned}$$

and by the right-hand thumb rule both are directed along $-\hat{\text k}$. Their sum is

$\vec{\text B}_1 = 2\left(\frac{\text µ_0}{4\text π}\frac{\text I}{\text R}\right)(-\hat{\text k})$

Field due to the semi-circular portion : The field at the centre due to a circular arc of radius R subtending an angle θ is

$\text B = \frac{\text µ_0}{4\text π}\frac{\text I}{\text R}\text θ$

For a semi-circle, θ = π, so

$$\begin{aligned}\vec{\text B}_2 = \frac{\text µ_0 \text I}{4\text R}(-\hat{\text i}) \\[1em] \vec{\text B}_2 = \text π\left(\frac{\text µ_0}{4\text π}\frac{\text I}{\text R}\right)(-\hat{\text i})\end{aligned}$$

Resultant field :

$$\begin{aligned}\vec{\text B} = \vec{\text B}_1 + \vec{\text B}_2 \\[1em] \vec{\text B} = -\frac{\text µ_0}{4\text π}\frac{\text I}{\text R}(\text π\hat{\text i} + 2\hat{\text k})\end{aligned}$$

$\vec{\text F}_1 = \vec{\text F}_2 = 0$

Reason — The magnetic field of a long solenoid is confined to its interior; outside the solenoid the field is zero.

The outer solenoid therefore lies in a region where the field of the inner solenoid is zero, so $\vec{\text F}_2 = 0$.

The inner solenoid lies inside the outer one, where the field is uniform and directed along the axis. The force on each current-element of a turn is radial, and on going round a complete turn these radial forces cancel in pairs. Hence the total force on the inner solenoid is also zero, $\vec{\text F}_1 = 0$.

$2\sin θ\sqrt{\frac{\text{πλgL}}{\text µ_0 \cos θ}}$

Reason — Given, threads of length L making an angle θ with the vertical, and wires of mass λ per unit length.

The separation between the two wires is x = 2L sin θ, and the currents are in opposite directions, so the force per unit length between them is repulsive,

$$\begin{aligned}\text F = \frac{\text µ_0}{2\text π}\frac{\text I^2}{2\text L \sin θ} \\[1em] \text F = \frac{\text µ_0 \text I^2}{4\text{πL} \sin θ}\end{aligned}$$

For equilibrium of each wire, resolving the tension T in the thread,

$\text T \sin θ = \text F, \qquad \text T \cos θ = \text{λg}$

Dividing,

$$\begin{aligned}\tan θ = \frac{\text F}{\text{λg}} \\[1em] \frac{\text µ_0 \text I^2}{4\text{πL} \sin θ} = \text{λg} \tan θ \\[1em] \text I^2 = \frac{4\text{πLλg} \sin^2 θ}{\text µ_0 \cos θ} \\[1em] \text I = 2 \sin θ\sqrt{\frac{\text{πλgL}}{\text µ_0 \cos θ}}\end{aligned}$$

(2) and (4) respectively

Reason — A current loop placed in a magnetic field is in stable equilibrium when its magnetic moment $\vec{\text M} \space$is parallel to $\vec{\text B} \space$, and in unstable equilibrium when $\vec{\text M} \space$is anti-parallel to $\vec{\text B} \space$.

The field is along the positive Z direction. Applying the right-hand palm rule no. 1 to each orientation, the magnetic moment of the loop in orientation (2) points along the positive Z direction, so it is in stable equilibrium, while in orientation (4) it points along the negative Z direction, so it is in unstable equilibrium.

1

Reason — Given, a wire of radius a carrying a steady current I distributed uniformly over the cross-section.

At the radial distance a/2 (inside the wire) :

$$\begin{aligned}\text B_1 = \frac{\text µ_0 \text I}{2\text π}\frac{(\text a/2)}{\text a^2} \\[1em] \text B_1 = \frac{\text µ_0 \text I}{4\text{πa}}\end{aligned}$$

At the radial distance 2a (outside the wire) :

$$\begin{aligned}\text B_2 = \frac{\text µ_0 \text I}{2\text π(2\text a)} \\[1em] \text B_2 = \frac{\text µ_0 \text I}{4\text{πa}} \\[1em] \frac{\text B_1}{\text B_2} = 1\end{aligned}$$

1

Reason — Given, a wire of radius a carrying a steady current I distributed uniformly across its cross-section.

Inside the wire, $\text B = \frac{\text µ_0 \text{Ir}}{2\text{πa}^2}$, so at r = a/2,

$$\begin{aligned}\text B_1 = \frac{\text µ_0 \text I}{2\text{πa}^2} \times \frac{\text a}{2} \\[1em] \text B_1 = \frac{\text µ_0 \text I}{4\text{πa}}\end{aligned}$$

Outside the wire, $\text B = \frac{\text µ_0 \text I}{2\text{πr}}$, so at r = 2a,

$$\begin{aligned}\text B_2 = \frac{\text µ_0 \text I}{4\text{πa}} \\[1em] \frac{\text B_1}{\text B_2} = 1\end{aligned}$$

n²B

Reason — Let the length of the wire be L.

For one turn : L = 2πr₁, so $\text r_1 = \frac{\text L}{2\text π}$ and

$$\begin{aligned}\text B = \frac{\text µ_0 \text I}{2\text r_1} \\[1em] \text B = \frac{\text µ_0 \text{πI}}{\text L}\end{aligned}$$

For n turns : L = n(2πr₂), so $\text r_2 = \frac{\text L}{2\text{πn}}$ and

$$\begin{aligned}\text B' = \frac{\text µ_0 \text n \text I}{2\text r_2} \\[1em] \text B' = \frac{\text µ_0 \text n \text I \times 2\text{πn}}{2\text L} \\[1em] \text B' = \frac{\text µ_0 \text{πn}^2 \text I}{\text L} \\[1em] \text B' = \text n^2 \text B\end{aligned}$$

$\frac{\text π^2}{8\sqrt{2}}$

Reason — Given, each wire has length l and carries the same current I.

Wire A, bent into a circle : l = 2πR, so $\text R = \frac{l}{2\text π}$ and

$$\begin{aligned}\text B_\text A = \frac{\text µ_0 \text I}{2\text R} \\[1em] \text B_\text A = \frac{\text µ_0 \text{πI}}{l}\end{aligned}$$

Wire B, bent into a square : the side is $\text a = \frac{l}{4}$, and the perpendicular distance of each side from the centre is $\frac{\text a}{2} = \frac{l}{8}$. For one side, φ₁ = φ₂ = 45°, so

$$\begin{aligned}\text B_{\text{side}} = \frac{\text µ_0}{4\text π}\frac{\text I}{(l/8)}(\sin 45° + \sin 45°) \\[1em] \text B_{\text{side}} = \frac{2\text µ_0 \text I}{\text π l} \times \sqrt{2}\end{aligned}$$

All the four sides give fields in the same direction, so

$$\begin{aligned}\text B_\text B = 4 \times \frac{2\sqrt{2}\text µ_0 \text I}{\text π l} \\[1em] \text B_\text B = \frac{8\sqrt{2}\text µ_0 \text I}{\text π l} \\[1em] \frac{\text B_\text A}{\text B_\text B} = \frac{\text µ_0 \text{πI}/l}{8\sqrt{2}\text µ_0 \text I/\text π l} \\[1em] \frac{\text B_\text A}{\text B_\text B} = \frac{\text π^2}{8\sqrt{2}}\end{aligned}$$

1 GHz

Reason — Given, B = 3.57 × 10^-2 T and $\frac{\text e}{\text m}$ = 1.76 × 10¹¹ C/kg.

The frequency of revolution of a charged particle in a magnetic field is

$$\begin{aligned}\text ν = \frac{\text{eB}}{2\text{πm}} \\[1em] \text ν = \frac{1}{2\text π}\left(\frac{\text e}{\text m}\right)\text B\end{aligned}$$

Substituting the values,

$$\begin{aligned}\text ν = \frac{(1.76 \times 10^{11})(3.57 \times 10^{-2})}{2 \times 3.14} \\[1em] \text ν = \frac{6.283 \times 10^9}{6.28} \\[1em] \text ν = 1 \times 10^9\ \text{Hz} = 1\ \text{GHz}\end{aligned}$$

$\frac{2\text µ_0 \text{Ii}}{3\text π}$

Reason — Given, a square loop ABCD of side L carrying a current i, placed coplanar with a long straight conductor XY carrying a current I, the nearer side being at a distance $\frac{\text L}{2}$ from XY.

The nearer side AB is at a distance $\frac{\text L}{2}$ and the farther side DC at a distance $\frac{\text L}{2} + \text L = \frac{3\text L}{2}$.

Force on the nearer side :

$$\begin{aligned}\text F_1 = \frac{\text µ_0}{2\text π}\frac{\text{Ii}}{(\text L/2)} \times \text L \\[1em] \text F_1 = \frac{\text µ_0 \text{Ii}}{\text π}\end{aligned}$$

Force on the farther side :

$$\begin{aligned}\text F_2 = \frac{\text µ_0}{2\text π}\frac{\text{Ii}}{(3\text L/2)} \times \text L \\[1em] \text F_2 = \frac{\text µ_0 \text{Ii}}{3\text π}\end{aligned}$$

The currents in these two sides are oppositely directed, so one force is an attraction and the other a repulsion. The forces on the other two sides are equal and opposite and cancel. Hence the net force is

$$\begin{aligned}\text F = \text F_1 - \text F_2 \\[1em] \text F = \frac{\text µ_0 \text{Ii}}{\text π} - \frac{\text µ_0 \text{Ii}}{3\text π} \\[1em] \text F = \frac{2\text µ_0 \text{Ii}}{3\text π}\end{aligned}$$

$\frac{\text µ_0 \text I}{4\text{πa}}6[\sqrt{3} - 1]$

Reason — Given, a symmetric star shaped loop carrying a current I, the distance between the diametrically opposite vertices being 4a.

The star is made up of 12 straight segments, and by symmetry each of them produces the same magnitude of field at the centre, all in the same direction.

For each segment the perpendicular distance from the centre is d = OA = a, and its ends subtend 60° and 30° at O on the same side of the normal OA (that is, φ₁ = 60°, φ₂ = −30°), so

$\text B_{\text{seg}} = \frac{\text µ_0}{4\text π}\frac{\text I}{\text d}(\sin \text φ_1 + \sin \text φ_2)$

Adding the contributions of all the twelve segments,

$$\begin{aligned}\text B = 12 \times \frac{\text µ_0}{4\text π}\frac{\text I}{\text a}[\sin 60° + \sin(-30°)] \\[1em] \text B = 12 \times \frac{\text µ_0 \text I}{4\text{πa}}\left(\frac{\sqrt{3}}{2} - \frac{1}{2}\right) \\[1em] \text B = \frac{\text µ_0 \text I}{4\text{πa}}6[\sqrt{3} - 1]\end{aligned}$$

$\frac{\text µ_0 \text I^2}{\sqrt{2}\text{πd}}$

Reason — Given, three parallel wires carrying the same current I along the same direction, the wire B being at a distance d from A and d from C, with AB perpendicular to BC.

The force per unit length on B due to A is

$\text F_1 = \frac{\text µ_0}{2\text π}\frac{\text I^2}{\text d}$

and that due to C is

$\text F_2 = \frac{\text µ_0}{2\text π}\frac{\text I^2}{\text d}$

Since all the currents are in the same direction, B is attracted towards A and towards C. The angle between BA and BC is 90°, so the two forces are at right angles and the resultant is

$$\begin{aligned}\text F = \sqrt{\text F_1^2 + \text F_2^2} \\[1em] \text F = \sqrt{2}\left(\frac{\text µ_0 \text I^2}{2\text{πd}}\right) \\[1em] \text F = \frac{\text µ_0 \text I^2}{\sqrt{2}\ \text{πd}}\end{aligned}$$

11.32 A

Reason — Given,

• mass per unit length, $\frac{\text m}{l}$ = 0.5 kg m^-1
• θ = 30°, B = 0.25 T, g = 9.8 m s^-2

The rod lies horizontally on the smooth inclined plane and the field is vertical, so the magnetic force IBl is horizontal. Balancing the components along the inclined plane,

$$\begin{aligned}\text{IB}l \cos θ = \text{mg} \sin θ \\[1em] \text I = \frac{\text m}{l} \times \frac{\text g \tan θ}{\text B} \\[1em] \text I = 0.5 \times \frac{(9.8)(\tan 30°)}{0.25} \\[1em] \text I = 0.5 \times \frac{(9.8)(0.5774)}{0.25} \\[1em] \text I = \frac{2.829}{0.25} \\[1em] \text I = 11.32\ \text A\end{aligned}$$

r_e < r_p = r_α

Reason — Given, the three particles have the same kinetic energy K in the same field B.

$\text r = \frac{\sqrt{2\text{mK}}}{\text{qB}} \quad \Rightarrow \quad \text r \propto \frac{\sqrt{\text m}}{\text q}$

Taking the proton mass as m and its charge as e,

• electron : $\text r_\text e \propto \frac{\sqrt{\text m_\text e}}{\text e}$, and m_e is very much smaller than m
• proton : $\text r_\text p \propto \frac{\sqrt{\text m}}{\text e}$
• α-particle : $\text r_α \propto \frac{\sqrt{4\text m}}{2\text e} = \frac{\sqrt{\text m}}{\text e}$

Hence r_p = r_α, and since m_e < m, r_e < r_p = r_α.

20 A, perpendicular into the page

Reason — Given, OP = OQ = 4 cm = 0.04 m and B = 10^-4 T at O.

The segments LP and QM lie along the x-axis, and O also lies on this axis. For these segments θ = 0 or 180°, so they develop no magnetic field at O.

The segments PS and QN are semi-infinite conductors, each at a perpendicular distance 0.04 m from O. Each therefore gives

$$\begin{aligned}\text B_1 = \frac{\text µ_0}{4\text π}\frac{\text I}{\text d}(\sin 0° + \sin 90°) \\[1em] \text B_1 = \frac{\text µ_0}{4\text π}\frac{\text I}{\text d}\end{aligned}$$

Applying the right-hand thumb rule, both these fields at O are directed into the page, so

$$\begin{aligned}\text B = 2 \times \frac{\text µ_0}{4\text π}\frac{\text I}{\text d} \\[1em] \text B = \frac{\text µ_0 \text I}{2\text{πd}} \\[1em] 10^{-4} = (2 \times 10^{-7}) \times \frac{\text I}{0.04} \\[1em] \text I = \frac{(10^{-4})(0.04)}{2 \times 10^{-7}} \\[1em] \text I = 20\ \text A\end{aligned}$$

zero

Reason — Given, two mutually perpendicular wires carrying equal currents I, and the point P at a perpendicular distance d from each of them.

The field at P due to each wire has the magnitude

$\text B_1 = \text B_2 = \frac{\text µ_0 \text I}{2\text{πd}}$

Applying the right-hand thumb rule to the two wires for the directions of current shown, the field due to one wire at P is directed into the plane of the page while that due to the other is directed out of the plane of the page. Being equal in magnitude and opposite in direction, they cancel,

$$\begin{aligned}\vec{\text B} = \vec{\text B}_1 + \vec{\text B}_2 \\[1em] \vec{\text B} = 0\end{aligned}$$

Reason — For a cylindrical conductor of radius R carrying a constant current distributed uniformly over its cross-section,

inside the conductor (d < R) :

$\text B = \frac{\text µ_0 \text I}{2\text π}\frac{\text d}{\text R^2} \quad \Rightarrow \quad \text B \propto \text d$

so B increases linearly with d, reaching its maximum value $\frac{\text µ_0 \text I}{2\text{πR}}$ at the surface;

outside the conductor (d > R) :

$\text B = \frac{\text µ_0 \text I}{2\text{πd}} \quad \Rightarrow \quad \text B \propto \frac{1}{\text d}$

so B decreases hyperbolically. This variation is represented by the first graph.

3 × 10^-5 C

Reason — Given,

• r = 10 cm = 0.1 m
• ω = 40π rad s^-1
• B = 3.8 × 10^-9 T

The rotating ring is equivalent to a current loop. If q is the charge on the ring, the equivalent current is

$$\begin{aligned}\text I = \frac{\text q}{\text T} \\[1em] \text I = \frac{\text{qω}}{2\text π} \\[1em] \text I = \frac{\text q(40\text π)}{2\text π} \\[1em] \text I = 20\text q\end{aligned}$$

The field at the centre of the ring is

$$\begin{aligned}\text B = \frac{\text µ_0 \text I}{2\text r} \\[1em] \text B = \frac{(4\text π \times 10^{-7})(20\text q)}{2 \times 0.1} \\[1em] 3.8 \times 10^{-9} = (1.2566 \times 10^{-4})\text q \\[1em] \text q = \frac{3.8 \times 10^{-9}}{1.2566 \times 10^{-4}} \\[1em] \text q = 3 \times 10^{-5}\ \text C\end{aligned}$$

$\frac{1}{\text N^2}$

Reason — Given, both the wires have the same length L and carry the same current.

For the single loop : L = 2πR_L, so $\text R_\text L = \frac{\text L}{2\text π}$ and

$$\begin{aligned}\text B_\text L = \frac{\text µ_0 \text I}{2\text R_\text L} \\[1em] \text B_\text L = \frac{\text µ_0 \text{πI}}{\text L}\end{aligned}$$

For the coil of N turns : L = N(2πR_C), so $\text R_\text C = \frac{\text L}{2\text{πN}}$ and

$$\begin{aligned}\text B_\text C = \frac{\text µ_0 \text N \text I}{2\text R_\text C} \\[1em] \text B_\text C = \frac{\text µ_0 \text{πN}^2 \text I}{\text L} \\[1em] \frac{\text B_\text L}{\text B_\text C} = \frac{1}{\text N^2}\end{aligned}$$

1.5 × 10^-5 T

Reason — Given, AB = 6 cm, PA = PB = 5 cm and I = 5 A.

The perpendicular distance of P from AB is

$$\begin{aligned}\text d = \sqrt{(5)^2 - (3)^2} \\[1em] \text d = \sqrt{25 - 9} \\[1em] \text d = 4\ \text{cm} = 0.04\ \text m\end{aligned}$$

For a straight conductor of finite length,

$\text B = \frac{\text µ_0}{4\text π}\frac{\text I}{\text d}(\sin \text φ_1 + \sin \text φ_2)$

Here, by symmetry, $\sin \text φ_1 = \sin \text φ_2 = \frac{3}{5} = 0.6$. Substituting,

$$\begin{aligned}\text B = (10^{-7}) \times \frac{5}{0.04} \times (0.6 + 0.6) \\[1em] \text B = (10^{-7})(125)(1.2) \\[1em] \text B = 1.5 \times 10^{-5}\ \text T\end{aligned}$$

5q

Reason — Given, $\vec{\text E} = 2\hat{\text i} + 3\hat{\text j}$ and the displacement from the origin to P(1, 1) is $\vec{\text d} = \hat{\text i} + \hat{\text j}$.

The magnetic force is always perpendicular to the velocity, so the work done by the magnetic field is zero. Only the electric force does work,

$$\begin{aligned}\text W = \text q\vec{\text E} \cdot \vec{\text d} \\[1em] \text W = \text q(2\hat{\text i} + 3\hat{\text j}) \cdot (\hat{\text i} + \hat{\text j}) \\[1em] \text W = \text q(2 + 3) \\[1em] \text W = 5\text q\end{aligned}$$

repulsive and equal to $\frac{\text µ_0 \text I_1 \text I_2}{4\text π}$

Reason — Given, a square loop of side a carrying a current I₂, the nearer side being at a distance a from the long wire carrying I₁.

Force on the nearer side (at a distance a) :

$$\begin{aligned}\text F_1 = \frac{\text µ_0}{2\text π}\frac{\text I_1 \text I_2}{\text a} \times \text a \\[1em] \text F_1 = \frac{\text µ_0 \text I_1 \text I_2}{2\text π}\end{aligned}$$

Force on the farther side (at a distance 2a) :

$$\begin{aligned}\text F_2 = \frac{\text µ_0}{2\text π}\frac{\text I_1 \text I_2}{2\text a} \times \text a \\[1em] \text F_2 = \frac{\text µ_0 \text I_1 \text I_2}{4\text π}\end{aligned}$$

The forces on the other two sides are equal and opposite and cancel out. The current in the nearer side is opposite to I₁, so F₁ is a repulsion, while F₂ is an attraction. Hence the net force is

$$\begin{aligned}\text F = \text F_1 - \text F_2 \\[1em] \text F = \frac{\text µ_0 \text I_1 \text I_2}{2\text π} - \frac{\text µ_0 \text I_1 \text I_2}{4\text π} \\[1em] \text F = \frac{\text µ_0 \text I_1 \text I_2}{4\text π}\end{aligned}$$

and it is repulsive, since the larger force is the repulsive one.

12.87 cm

Reason — Given, energy = 100 eV, B = 1.5 × 10^-3 T, the field extending over x = 2 cm, and the screen 8 cm from S.

Since $\vec{\text B} \space$is along $+\hat{\text k}$, the field points out of the page (the crosses in the question figure do not match the stated $\hat{\text k}$), and the force on the electron, $-\text e(\vec{\text v} \times \vec{\text B})$, is along +y, so it bends towards Q.

Speed of the electron :

$$\begin{aligned}\text v = \sqrt{\frac{2\text K}{\text m}} \\[1em] \text v = \sqrt{\frac{2(100)(1.6 \times 10^{-19})}{9.1 \times 10^{-31}}} \\[1em] \text v = 5.93 \times 10^6\ \text{ms}^{-1}\end{aligned}$$

Radius of the circular path inside the field :

$$\begin{aligned}\text r = \frac{\text{mv}}{\text{eB}} \\[1em] \text r = \frac{(9.1 \times 10^{-31})(5.93 \times 10^6)}{(1.6 \times 10^{-19})(1.5 \times 10^{-3})} \\[1em] \text r = 2.25\ \text{cm}\end{aligned}$$

Deflection inside the field region : with x = 2 cm,

$$\begin{aligned}\text y_1 = \text r - \sqrt{\text r^2 - \text x^2} \\[1em] \text y_1 = 2.25 - \sqrt{(2.25)^2 - (2)^2} \\[1em] \text y_1 = 2.25 - 1.03 \\[1em] \text y_1 = 1.22\ \text{cm}\end{aligned}$$

Deflection outside the field region : the electron leaves the field at an angle θ with

$$\begin{aligned}\sin θ = \frac{\text x}{\text r} \\[1em] \sin θ = \frac{2}{2.25} \\[1em] \sin θ = 0.889 \\[1em] \cos θ = 0.458 \\[1em] \tan θ = 1.94\end{aligned}$$

and then travels in a straight line over the remaining 8 − 2 = 6 cm,

$$\begin{aligned}\text y_2 = 6 \tan θ \\[1em] \text y_2 = 6 \times 1.94 \\[1em] \text y_2 = 11.64\ \text{cm}\end{aligned}$$

Total distance :

$$\begin{aligned}\text d = \text y_1 + \text y_2 \\[1em] \text d = 1.22 + 11.64 \\[1em] \text d = 12.86\ \text{cm}\end{aligned}$$

This agrees with option 2 (12.87 cm) to within rounding; with the unrounded radius (2.248 cm), d ≈ 12.9 cm.

$\text x = \pm\frac{\text I_1 \text d}{(\text I_1 - \text I_2)}$

Reason — Given, the wires A and B carry currents I₁ and I₂ in opposite directions, separated by a distance d, and the wire C is at a distance x from A.

The net force on C is zero when the two forces per unit length are equal and opposite. Since the currents in A and B are oppositely directed, the two forces act in the same direction at every point between the wires, so no null point can lie between them; C must lie outside.

C beyond B (distance x − d from B) :

$\frac{\text µ_0}{2\text π}\frac{\text I_1 \text I}{\text x} = \frac{\text µ_0}{2\text π}\frac{\text I_2 \text I}{(\text x - \text d)}$

Solving,

$$\begin{aligned}\text I_1(\text x - \text d) = \text I_2 \text x \\[1em] \text x(\text I_1 - \text I_2) = \text I_1 \text d \\[1em] \text x = \frac{\text I_1 \text d}{\text I_1 - \text I_2}\end{aligned}$$

C beyond A (distance x + d from B) :

$$\begin{aligned}\frac{\text I_1}{\text x} = \frac{\text I_2}{\text x + \text d} \\[1em] \text x = \frac{\text I_1 \text d}{\text I_2 - \text I_1} \\[1em] \text x = -\frac{\text I_1 \text d}{\text I_1 - \text I_2}\end{aligned}$$

Combining the two cases,

$\text x = \pm\frac{\text I_1 \text d}{(\text I_1 - \text I_2)}$

2 : 1

Reason — Given, the ionized hydrogen atoms and the α-particles have the same momentum p and enter the same field B perpendicularly.

$\text r = \frac{\text p}{\text{qB}} \quad \Rightarrow \quad \text r \propto \frac{1}{\text q}$

The charge on an ionized hydrogen atom is e and that on an α-particle is 2e, so

$$\begin{aligned}\frac{\text r_\text H}{\text r_α} = \frac{\text q_α}{\text q_\text H} \\[1em] \frac{\text r_\text H}{\text r_α} = \frac{2\text e}{\text e} \\[1em] \frac{\text r_\text H}{\text r_α} = \frac{2}{1}\end{aligned}$$

r_e < r_p = r_He

Reason — Given, the three particles have the same energy K in the same magnetic field B.

$\text r = \frac{\sqrt{2\text{mK}}}{\text{qB}} \quad \Rightarrow \quad \text r \propto \frac{\sqrt{\text m}}{\text q}$

Taking the proton mass as m and its charge as e,

• proton : $\text r_\text p \propto \frac{\sqrt{\text m}}{\text e}$
• helium nucleus : m_He = 4m and q_He = 2e, so $\text r_{\text{He}} \propto \frac{\sqrt{4\text m}}{2\text e} = \frac{\sqrt{\text m}}{\text e}$
• electron : $\text r_\text e \propto \frac{\sqrt{\text m_\text e}}{\text e}$, and m_e is very much smaller than m

Hence r_p = r_He, while r_e is much smaller, giving r_e < r_p = r_He.

2.0 × 10^-24 kg

Reason — Given,

• r = 0.5 cm = 5 × 10^-3 m
• B = 0.5 T
• E = 100 V/m
• q = 1.6 × 10^-19 C

The particle moves in a straight path when the electric force balances the magnetic force,

$$\begin{aligned}\text{qE} = \text{qvB} \\[1em] \text v = \frac{\text E}{\text B} \\[1em] \text v = \frac{100}{0.5} \\[1em] \text v = 200\ \text{ms}^{-1}\end{aligned}$$

From the radius of the circular path, $\text r = \frac{\text{mv}}{\text{qB}}$,

$$\begin{aligned}\text m = \frac{\text{qBr}}{\text v} \\[1em] \text m = \frac{(1.6 \times 10^{-19})(0.5)(5 \times 10^{-3})}{200} \\[1em] \text m = \frac{4 \times 10^{-22}}{200} \\[1em] \text m = 2.0 \times 10^{-24}\ \text{kg}\end{aligned}$$

$\frac{\text{qdB}_0}{\text m}$

Reason — Given, $\vec{\text v} = -\text v\hat{\text i}$, $\vec{\text B} = \text B_0\hat{\text k}$ and the screen at a distance d in the Y-Z plane.

The velocity is perpendicular to the field, so the particle describes a circle of radius

$\text r = \frac{\text{mv}}{\text{qB}_0}$

Starting along the −X direction, the greatest distance the particle can penetrate along its initial direction of motion is equal to the radius r of the circle, after which it turns back.

The particle will not hit the screen so long as

$$\begin{aligned}\text r \lt \text d \quad \Rightarrow \quad \frac{\text{mv}}{\text{qB}_0} \lt \text d \\[1em] \text v \lt \frac{\text{qdB}_0}{\text m}\end{aligned}$$

Hence the maximum value of v is $\frac{\text{qdB}_0}{\text m}$.

0.44 m

Reason — Given, B = 0.3 T, v = 4 × 10⁵ ms^-1, θ = 60°, m = 1.67 × 10^-27 kg and q = 1.6 × 10^-19 C.

The pitch of the helical path is the distance advanced along the field in one complete revolution,

$$\begin{aligned}\text P = \text v \cos θ \times \frac{2\text{πm}}{\text{qB}} \\[1em] \text P = (4 \times 10^5)(\cos 60°) \times \frac{2 \times 3.14 \times 1.67 \times 10^{-27}}{(1.6 \times 10^{-19})(0.3)} \\[1em] \text P = (2 \times 10^5) \times \frac{1.049 \times 10^{-26}}{4.8 \times 10^{-20}} \\[1em] \text P = 0.0437\ \text m\end{aligned}$$

The proton completes 10 revolutions while crossing the region, so

$$\begin{aligned}l = 10 \times \text P \\[1em] l = 10 \times 0.0437 \\[1em] l = 0.44\ \text m\end{aligned}$$

4 cm

Reason — Given, v = 4 × 10⁵ ms^-1, B = 0.3 T, θ = 60°, m = 1.67 × 10^-27 kg and q = 1.69 × 10^-19 C.

The pitch of the helical path is

$\text P = \text v \cos θ \times \frac{2\text{πm}}{\text{qB}}$

Substituting the values,

$$\begin{aligned}\text P = (4 \times 10^5)(0.5) \times \frac{2 \times 3.14 \times 1.67 \times 10^{-27}}{(1.69 \times 10^{-19})(0.3)} \\[1em] \text P = (2 \times 10^5) \times \frac{1.049 \times 10^{-26}}{5.07 \times 10^{-20}} \\[1em] \text P = 0.0414\ \text m \\[1em] \text P \approx 4\ \text{cm}\end{aligned}$$

5 × 10^-3 T, along +Z direction

Reason — Given, v = 6 × 10⁶ ms^-1 along +X, and E = 300 V/cm = 3 × 10⁴ V/m along +Y.

The electron keeps moving along X, so the electric and magnetic forces on it balance,

$$\begin{aligned}\text{eE} = \text{evB} \\[1em] \text B = \frac{\text E}{\text v} \\[1em] \text B = \frac{3 \times 10^4}{6 \times 10^6} \\[1em] \text B = 5 \times 10^{-3}\ \text T\end{aligned}$$

Direction : The electron carries a charge −e, so the electric force on it is along −Y. The magnetic force must therefore be along +Y. Taking $\vec{\text B} = \text B\hat{\text k}$,

$$\begin{aligned}\vec{\text F} = (-\text e)(\text v\hat{\text i} \times \text B\hat{\text k}) \\[1em] \vec{\text F} = (-\text e)(-\text{vB}\hat{\text j}) \\[1em] \vec{\text F} = \text{evB}\hat{\text j}\end{aligned}$$

which is along +Y, as required. Hence the field is directed along the +Z direction.

8 × 10^-20 N

Reason — Given, I = 5 A, v = 10⁵ m/s and r = 20 cm = 0.2 m.

The field produced by the conductor at the position of the electron is

$$\begin{aligned}\text B = \frac{\text µ_0}{4\text π}\frac{2\text I}{\text r} \\[1em] \text B = (10^{-7}) \times \frac{2 \times 5}{0.2} \\[1em] \text B = 5 \times 10^{-6}\ \text T\end{aligned}$$

The electron moves parallel to the conductor, so it moves at right angles to this field and

$$\begin{aligned}\text F = \text{evB} \\[1em] \text F = (1.6 \times 10^{-19})(10^5)(5 \times 10^{-6}) \\[1em] \text F = 8 \times 10^{-20}\ \text N\end{aligned}$$

Reason — For a thick cable of radius R carrying a current I uniformly distributed over its cross-section,

inside the cable (r < R) : the current enclosed by a circle of radius r is $\text I\frac{\text r^2}{\text R^2}$, so

$\text B = \frac{\text µ_0 \text{Ir}}{2\text{πR}^2} \quad \Rightarrow \quad \text B \propto \text r$

outside the cable (r > R) : the whole current is enclosed, so

$\text B = \frac{\text µ_0 \text I}{2\text{πr}} \quad \Rightarrow \quad \text B \propto \frac{1}{\text r}$

Hence B increases linearly with r up to the surface of the cable, where it reaches its maximum value, and then falls off hyperbolically. This variation is represented by the fourth graph.

$-6\hat{\text i} - 6\hat{\text j} - 8\hat{\text k}$

Reason — Given, q = 1, $\vec{\text v} = 2\hat{\text i} + 4\hat{\text j} + 6\hat{\text k}$, $\vec{\text B} = \text B\hat{\text i} + \text B\hat{\text j} + \text B_0\hat{\text k}$ and $\vec{\text F} = 4\hat{\text i} - 20\hat{\text j} + 12\hat{\text k}$.

Evaluating the vector product,

$$\vec{\text v} \times \vec{\text B} = \begin{vmatrix} \hat{\text i} & \hat{\text j} & \hat{\text k} \\ 2 & 4 & 6 \\ \text B & \text B & \text B_0 \end{vmatrix} \\[1em] = \hat{\text i}(4\text B_0 - 6\text B) - \hat{\text j}(2\text B_0 - 6\text B) + \hat{\text k}(2\text B - 4\text B)$$

Comparing with $\vec{\text F} \space$, from the $\hat{\text k}$ component,

$$\begin{aligned}-2\text B = 12 \\[1em] \text B = -6\end{aligned}$$

From the $\hat{\text i}$ component,

$$\begin{aligned}4\text B_0 - 6(-6) = 4 \\[1em] 4\text B_0 = 4 - 36 \\[1em] 4\text B_0 = -32 \\[1em] \text B_0 = -8\end{aligned}$$

Checking with the $\hat{\text j}$ component, $-[2(-8) - 6(-6)] = -[-16 + 36] = -20$, which agrees.

$\vec{\text B} = -6\hat{\text i} - 6\hat{\text j} - 8\hat{\text k}$

$\frac{\text µ_0 \text I}{4\text{πxy}}[\sqrt{\text x^2 + \text y^2} + (\text x + \text y)]$

Reason — Given, two semi-infinite conductors along the X and the Y axes with their common end at the origin, carrying equal currents I, and P the point (x, y).

For a straight conductor of finite length,

$\text B = \frac{\text µ_0}{4\text π}\frac{\text I}{\text d}(\sin \text φ_1 + \sin \text φ_2)$

Due to the conductor along the X-axis : the perpendicular distance of P is y; the angles are 90° (end at infinity) and θ₁, with $\sin \text θ_1 = \frac{\text x}{\sqrt{\text x^2 + \text y^2}}$. Hence

$\text B_1 = \frac{\text µ_0 \text I}{4\text{πy}}\left[1 + \frac{\text x}{\sqrt{\text x^2 + \text y^2}}\right]$

Due to the conductor along the Y-axis : the perpendicular distance of P is x; the angles are 90° (end at infinity) and θ₂, with $\sin \text θ_2 = \frac{\text y}{\sqrt{\text x^2 + \text y^2}}$. Hence

$\text B_2 = \frac{\text µ_0 \text I}{4\text{πx}}\left[1 + \frac{\text y}{\sqrt{\text x^2 + \text y^2}}\right]$

Both the fields at P are in the same direction, so writing $\text R = \sqrt{\text x^2 + \text y^2}$ and adding,

$$\begin{aligned}\text B = \frac{\text µ_0 \text I}{4\text{πxy}}\left[\text x + \frac{\text x^2}{\text R} + \text y + \frac{\text y^2}{\text R}\right] \\[1em] \text B = \frac{\text µ_0 \text I}{4\text{πxy}}\left[(\text x + \text y) + \frac{\text x^2 + \text y^2}{\text R}\right] \\[1em] \text B = \frac{\text µ_0 \text I}{4\text{πxy}}\left[\sqrt{\text x^2 + \text y^2} + (\text x + \text y)\right]\end{aligned}$$

$\frac{\text µ_0 \text{IN}}{2(\text b - \text a)}\ln\left(\frac{\text b}{\text a}\right)$

Reason — Given, a spiral of N turns wound tightly between the inner radius a and the outer radius b, carrying a current I.

The N turns are spread uniformly over the radial width (b − a), so the number of turns lying between the radii r and r + dr is

$\text{dN} = \frac{\text N}{\text b - \text a}\text{dr}$

Each of these turns behaves as a circular coil of radius r, so the field it produces at the centre is

$$\begin{aligned}\text{dB} = \frac{\text µ_0 \text I}{2\text r}\text{dN} \\[1em] \text{dB} = \frac{\text µ_0 \text{IN}}{2(\text b - \text a)}\frac{\text{dr}}{\text r}\end{aligned}$$

Integrating from r = a to r = b,

$$\begin{aligned}\text B = \frac{\text µ_0 \text{IN}}{2(\text b - \text a)}\int_\text a^\text b \frac{\text{dr}}{\text r} \\[1em] \text B = \frac{\text µ_0 \text{IN}}{2(\text b - \text a)}\left[\ln \text r\right]_\text a^\text b \\[1em] \text B = \frac{\text µ_0 \text{IN}}{2(\text b - \text a)}(\ln \text b - \ln \text a) \\[1em] \text B = \frac{\text µ_0 \text{IN}}{2(\text b - \text a)}\ln\left(\frac{\text b}{\text a}\right)\end{aligned}$$

$\frac{3}{2}\frac{\text r^2}{\text a^2}$

Reason — Given, a coil of radius a, and a point on its axis at a distance r from the centre, with r < a.

At the centre :

$\text B_\text c = \frac{\text µ_0 \text I}{2\text a}$

On the axis at a distance r :

$\text B_\text{axis} = \frac{\text µ_0 \text I \text a^2}{2(\text a^2 + \text r^2)^{3/2}}$

Taking the ratio,

$$\begin{aligned}\frac{\text B_\text{axis}}{\text B_\text c} = \frac{\text a^3}{(\text a^2 + \text r^2)^{3/2}} \\[1em] \frac{\text B_\text{axis}}{\text B_\text c} = \left(1 + \frac{\text r^2}{\text a^2}\right)^{-3/2}\end{aligned}$$

Since r < a, the term $\frac{\text r^2}{\text a^2}$ is small, and expanding by the binomial theorem and retaining only the first order term,

$\frac{\text B_\text{axis}}{\text B_\text c} \approx 1 - \frac{3}{2}\frac{\text r^2}{\text a^2}$

Hence the fractional change is

$\frac{\text B_\text c - \text B_\text{axis}}{\text B_\text c} = \frac{3}{2}\frac{\text r^2}{\text a^2}$

Reason — For a wire of radius R carrying a current I distributed uniformly over the cross-section,

$\text B = \frac{\text µ_0 \text{Ir}}{2\text{πR}^2}\ (\text r \lt \text R), \qquad \text B = \frac{\text µ_0 \text I}{2\text{πr}}\ (\text r \gt \text R)$

So each curve rises linearly up to the surface of its own wire and then falls off as $\frac{1}{\text r}$. The maximum value occurs at the surface and is

$\text B_\text{max} = \frac{\text µ_0 \text I}{2\text{πR}}$

Since a < b, the wire a has its peak at a smaller radius and, being inversely proportional to R, a higher peak, while the wire b peaks later and lower. For r greater than b both the wires carry the same enclosed current I, so the two curves coincide beyond that point. The option graphs draw the two tails separately only schematically; the correct choice is the one in which a peaks earlier and higher than b.

This behaviour is represented by the third graph.

Note: In the question figure, all four option graphs draw the curves for a and b as two separate tails beyond r = b. Since both wires enclose the full current I there, $\text B = \frac{\text µ_0 \text I}{2\text{πr}}$ for both, so the two curves should coincide.

A-III, B-II, C-IV, D-I

Reason — A. Capacitance, C : $\text C = \frac{\text Q}{\text V}$, so its dimensions are $\frac{[\text{AT}]}{[\text{ML}^2\text T^{-3}\text A^{-1}]}$ = M^-1 L^-2 T⁴ A², which is III.

B. Permittivity of free space, ε₀ : from $\text F = \frac{1}{4\text{πε}_0}\frac{\text q_1 \text q_2}{\text r^2}$, the dimensions are $\frac{[\text q_1][\text q_2]}{[\text F][\text r^2]} = \frac{[\text A^2\text T^2]}{[\text{MLT}^{-2}][\text L^2]}$ = M^-1 L^-3 T⁴ A², which is II.

C. Permeability of free space, µ₀ : the unit of µ₀ is newton/ampere², so

$$\begin{aligned}\text{Dimensions of } \text µ_0 = \frac{[\text M \text L \text T^{-2}]}{[\text A^2]} \\[1em] \text{Dimensions of } \text µ_0 = [\text M \text L \text T^{-2} \text A^{-2}]\end{aligned}$$

which is IV.

D. Electric field, E : $\text E = \frac{\text F}{\text q}$ has the dimensions $\frac{[\text{MLT}^{-2}]}{[\text{AT}]}$ = M L T^-3 A^-1, which is I.

lighter ion will be deflected more than heavier ion

Reason — Given, m₁ = 4 amu with q₁ = 2e, and m₂ = 16 amu with q₂ = 3e, both having the same kinetic energy K.

$\text r = \frac{\sqrt{2\text{mK}}}{\text{qB}} \quad \Rightarrow \quad \text r \propto \frac{\sqrt{\text m}}{\text q}$

For the lighter ion :

$\text r_1 \propto \frac{\sqrt{4}}{2\text e} = \frac{1}{\text e}$

For the heavier ion :

$\text r_2 \propto \frac{\sqrt{16}}{3\text e} = \frac{4}{3\text e}$

So r₁ < r₂. The smaller the radius of the circular path, the sharper is the bending, so the lighter ion is deflected more than the heavier ion.

Maximum current through R is 1.2 × 10^-6 A

Maximum charge on capacitor C₀ is 8.4 × 10^-12 C

Reason — Given, I = 2 A, v = 3.0 m/s, R = 1.4 Ω, C₀ = 5.0 µF, the ends of the rod being at 1 cm and 4 cm from the wire, and ln 2 = 0.7.

The rod moves in the non-uniform field of the wire, so the emf is obtained by integrating over its length. At a distance x from the wire the field is $\text B = \frac{\text µ_0 \text I}{2\text{πx}}$, and the element dx contributes Bv dx. Hence

$$\begin{aligned}\text e = \int_{0.01}^{0.04} \frac{\text µ_0 \text I}{2\text{πx}}\text v\ \text{dx} \\[1em] \text e = \frac{\text µ_0 \text{Iv}}{2\text π}\left[\ln \text x\right]_{0.01}^{0.04} \\[1em] \text e = \frac{\text µ_0 \text{Iv}}{2\text π}\ln\left(\frac{0.04}{0.01}\right) \\[1em] \text e = (2 \times 10^{-7})(2)(3.0)\ln 4\end{aligned}$$

Using ln 4 = 2 ln 2 = 1.4,

$$\begin{aligned}\text e = (1.2 \times 10^{-6})(1.4) \\[1em] \text e = 1.68 \times 10^{-6}\ \text V\end{aligned}$$

Maximum current flows at t = 0, when the capacitor is still uncharged,

$$\begin{aligned}\text I_\text{max} = \frac{\text e}{\text R} \\[1em] \text I_\text{max} = \frac{1.68 \times 10^{-6}}{1.4} \\[1em] \text I_\text{max} = 1.2 \times 10^{-6}\ \text A\end{aligned}$$

Maximum charge is reached when the capacitor is fully charged and the current stops,

$$\begin{aligned}\text Q_\text{max} = \text C_0 \text e \\[1em] \text Q_\text{max} = (5.0 \times 10^{-6})(1.68 \times 10^{-6}) \\[1em] \text Q_\text{max} = 8.4 \times 10^{-12}\ \text C\end{aligned}$$

Hence the first and the third statements are correct.

B ∝ r

Reason — Given, a wire of radius R carrying a current I distributed uniformly over its cross-section, and a point at a distance r with r < R.

Applying Ampere's circuital law to a circle of radius r drawn coaxially inside the wire, the current enclosed is

$$\begin{aligned}\text I_{\text{enc}} = \text I \times \frac{\text{πr}^2}{\text{πR}^2} \\[1em] \text I_{\text{enc}} = \frac{\text{Ir}^2}{\text R^2} \\[1em] \text B(2\text{πr}) = \text µ_0 \frac{\text{Ir}^2}{\text R^2} \\[1em] \text B = \frac{\text µ_0 \text{Ir}}{2\text{πR}^2}\end{aligned}$$

Since µ₀, I and R are constants, B ∝ r.

Reason — For a hollow conducting cylinder of radius R carrying a uniform current along its surface,

inside the cylinder (r < R) : the circle of radius r encloses no current, so by Ampere's circuital law

$$\begin{aligned}\text B(2\text{πr}) = \text µ_0 (0) \\[1em] \text B = 0\end{aligned}$$

outside the cylinder (r > R) : the whole current I is enclosed, so

$\text B = \frac{\text µ_0 \text I}{2\text{πr}} \quad \Rightarrow \quad \text B \propto \frac{1}{\text r}$

Hence B is zero everywhere inside, rises abruptly to $\frac{\text µ_0 \text I}{2\text{πR}}$ at the surface, and then falls off hyperbolically. This is represented by the fourth graph.

$\frac{\text x^2}{\text a^2}$

Reason — Given, the inner wire of radius a carrying a current I₀ uniformly distributed over its cross-section.

(i) For x < a : the circle of radius x encloses only a part of the current,

$$\begin{aligned}\text I_{\text{enc}} = \text I_0\frac{\text x^2}{\text a^2} \\[1em] \text B_1 = \frac{\text µ_0 \text I_0 \text x}{2\text{πa}^2}\end{aligned}$$

(ii) For a < x < b : the whole current of the inner wire is enclosed and the outer shell is not yet reached, so

$\text B_2 = \frac{\text µ_0 \text I_0}{2\text{πx}}$

Taking the ratio,

$$\begin{aligned}\frac{\text B_1}{\text B_2} = \frac{\text µ_0 \text I_0 \text x}{2\text{πa}^2} \times \frac{2\text{πx}}{\text µ_0 \text I_0} \\[1em] \frac{\text B_1}{\text B_2} = \frac{\text x^2}{\text a^2}\end{aligned}$$

6

Reason — Given, $\vec{\text B} = 2\hat{\text i} + 3\hat{\text j}$ and $\vec{\text a} = α\hat{\text i} - 4\hat{\text j}$.

The magnetic force, and hence the acceleration, is always perpendicular to the magnetic field, since

$\vec{\text F} = \text q(\vec{\text v} \times \vec{\text B})$

is perpendicular to $\vec{\text B} \space$. Therefore

$$\begin{aligned}\vec{\text a} \cdot \vec{\text B} = 0 \\[1em] (α\hat{\text i} - 4\hat{\text j}) \cdot (2\hat{\text i} + 3\hat{\text j}) = 0 \\[1em] 2α - 12 = 0 \\[1em] α = 6\end{aligned}$$

192 kV m^-1

Reason — Given, energy of the electron = 728 eV, B = 12 mT = 12 × 10^-3 T and m = 9.1 × 10^-31 kg.

Speed of the electron :

$$\begin{aligned}\text v = \sqrt{\frac{2\text K}{\text m}} \\[1em] \text v = \sqrt{\frac{2(728)(1.6 \times 10^{-19})}{9.1 \times 10^{-31}}} \\[1em] \text v = \sqrt{2.56 \times 10^{14}} \\[1em] \text v = 1.6 \times 10^7\ \text{ms}^{-1}\end{aligned}$$

Condition for the electron to pass undeflected : the electric force balances the magnetic force,

$$\begin{aligned}\text{eE} = \text{evB} \\[1em] \text E = \text{vB} \\[1em] \text E = (1.6 \times 10^7)(12 \times 10^{-3}) \\[1em] \text E = 1.92 \times 10^5\ \text{V m}^{-1} \\[1em] \text E = 192\ \text{kV m}^{-1}\end{aligned}$$

Both (A) and (R) are true and (R) is the correct explanation of (A).

Reason — Assertion (A) is correct : In a uniform magnetic field the speed of a moving charged particle, and hence its kinetic energy, remains unchanged.

Reason (R) is also correct : The magnetic force $\vec{\text F} = \text q(\vec{\text v} \times \vec{\text B})$ is always perpendicular to the direction of motion of the particle.

Because the force is perpendicular to the displacement, the work done on the particle is zero, and so its speed and energy cannot change. Hence the Reason correctly explains the Assertion.

Statement I is correct but Statement II is incorrect.

Reason — Statement I is correct : The electric force qE acts along the direction of the field and does work on the charged particle, so it changes the speed and hence the kinetic energy. The magnetic force is perpendicular to the velocity, does no work, and therefore does not change the kinetic energy.

Statement II is incorrect : The electric force accelerates a positively charged particle along the direction of the electric field, not perpendicular to it. The magnetic force accelerates the moving charged particle perpendicular to the direction of the magnetic field, not along it. Both parts of the statement are wrongly stated.

30 A

Reason — Given,

• linear density, $\frac{\text m}{l}$ = 0.45 kg m^-1
• θ = 45°, B = 0.15 T, g = 10 m/s²

The rod rests on a smooth inclined plane and the field is vertically upward, so the magnetic force IBl on the rod is horizontal. Balancing the components along the inclined plane,

$$\begin{aligned}\text{IB}l \cos θ = \text{mg} \sin θ \\[1em] \text I = \frac{\text m}{l} \times \frac{\text g \tan θ}{\text B} \\[1em] \text I = 0.45 \times \frac{(10)(\tan 45°)}{0.15} \\[1em] \text I = \frac{4.5}{0.15} \\[1em] \text I = 30\ \text A\end{aligned}$$