JEE Physics Moving Charges and Magnetism MCQs Set 18

Answer: (a) \( 0.5\ \text{A m}^2 \)

Explanation:
1. Each turn has moment \( iS \) with \( S=\frac{\pi d^2}{4} \), directed along the local tangent of the toroid.
2. Turns at angle \( \varphi \) from one end contribute the component \( \sin\varphi \) along the symmetry line; summing over \( N\,d\varphi/\pi \) turns over half a circle gives \( m=NiS\cdot\frac2\pi=\frac{Nid^2}{2} \).
3. Numerically: \( \frac{500\times0.8\times(0.05)^2}{2}=0.5 \) A m\( ^2 \).

Teacher's Note:
1. Draw the directions of the fields and forces first; then use magnitudes for the numbers.

Answer: (c) \( \frac{\mu_0}{4\pi}\cdot\frac{4I}{a} \) along +y

Explanation:
1. Each wire is at distance \( a\sqrt2 \) and produces \( \frac{\mu_0I}{2\pi a\sqrt2} \), perpendicular to its line to the centre.
2. Components along x cancel in pairs; the y-components of all four add (checked with vector sums).
3. Total \( 4\times\frac{\mu_0I}{2\pi a\sqrt2}\cdot\frac1{\sqrt2}=\frac{\mu_0I}{\pi a}=\frac{\mu_0}{4\pi}\frac{4I}{a} \), along +y.

Teacher's Note:
1. Draw the directions of the fields and forces first; then use magnitudes for the numbers.

Answer: (a) \( 9\times10^{9} \)

Explanation:
1. On the axis \( E=\frac{Qx}{4\pi\varepsilon_0(R^2+x^2)^{3/2}} \) and \( B=\frac{\mu_0QfR^2}{2(R^2+x^2)^{3/2}} \).
2. Energy density ratio \( \frac{\varepsilon_0E^2/2}{B^2/2\mu_0}=\left(\frac{E}{cB}\right)^2 \) with \( \frac{E}{B}=\frac{xc^2}{2\pi fR^2} \).
3. So ratio \( =\left(\frac{xc}{2\pi fR^2}\right)^2=\frac{(0.2\times3\times10^8)^2}{4\times10\times10^8\times10^{-4}}=9\times10^9 \).

Teacher's Note:
1. Draw the directions of the fields and forces first; then use magnitudes for the numbers.

Answer: (c) \( \frac{\pi m}{qB} \)

Explanation:
1. Momentum is conserved, so the particles start with equal and opposite velocities; the radii \( \frac{mv}{qB} \) are equal.
2. Opposite charges and opposite velocities mean that both are pushed to the same side, so they move on the same circle in opposite directions.
3. They meet at the diametrically opposite point after half a period: \( t=\frac T2=\frac{\pi m}{qB} \).

Teacher's Note:
1. Draw the directions of the fields and forces first; then use magnitudes for the numbers.

Answer: (a) 48

Explanation:
1. Momenta are equal and opposite, charges are equal in magnitude, so the radii are equal and both travel on the same circle, in opposite senses.
2. Angular speeds: \( \omega=\frac{qB}{M} \): the ratio is \( \frac{\omega_A}{\omega_P}=\frac{96}{4}=24 \) (alpha: mass number 4, ion: 96).
3. They meet when the angles add to \( 2\pi \): \( \theta_A\left(1+\frac1{24}\right)=2\pi \), so \( \theta_A=\frac{48\pi}{25} \) and n = 48.

Teacher's Note:
1. Draw the directions of the fields and forces first; then use magnitudes for the numbers.

Answer: (c) \( \vec B=-\hat k \) T and zero force

Explanation:
1. With \( \vec F=-e\,\vec v\times\vec B \): from \( \hat i\times\vec B=\hat j \) we get \( B_y=0,\ B_z=-1 \).
2. From the second condition \( (\hat i-\hat j)\times\vec B=\hat i+\hat j \) we get \( B_x=0 \) and \( B_z=-1 \) again: consistent, so \( \vec B=-\hat k \) T.
3. Then \( \vec v_3=\hat i\times(\hat i-\hat j)=-\hat k \) is parallel to B, so the force is zero.

Teacher's Note:
1. Draw the directions of the fields and forces first; then use magnitudes for the numbers.

Answer: (a) \( \frac{\mu_0i_0i}{2\pi}\ln\left(1+\frac\ell a\right) \) in the direction of \( i_0 \)

Explanation:
1. An element dx at distance x has field \( \frac{\mu_0i_0}{2\pi x} \) perpendicular to the plane, so the force is \( i\,dx\frac{\mu_0i_0}{2\pi x} \), all in the same direction. Using the right-hand rule, the field at AB is into the page and the force on a current directed away from the wire is along \( i_0 \).
2. Integrating from \( x=a \) to \( a+\ell \): \( F=\frac{\mu_0i_0i}{2\pi}\ln\frac{a+\ell}{a} \).

Teacher's Note:
1. Draw the directions of the fields and forces first; then use magnitudes for the numbers.

Answer: (c) \( 2\times10^{-5}\ln\frac32 \) N

Explanation:
1. The central wire's field at distance r is \( \frac{\mu_0i}{2\pi r} \), perpendicular to the radial segment (and horizontal), so the force on the segment is \( F=\int_{r_1}^{r_2}\frac{\mu_0i^2}{2\pi r}dr=\frac{\mu_0i^2}{2\pi}\ln\frac{r_2}{r_1} \).
2. Numerically \( 2\times10^{-7}\times100\times\ln1.5=2\times10^{-5}\ln\frac32 \) N (about \( 8.1\times10^{-6} \) N).

Teacher's Note:
1. Draw the directions of the fields and forces first; then use magnitudes for the numbers.

Answer: (a) \( \frac{E_0m}{qB_0^2}\left[1-\cos\frac{qB_0t}{m}\right] \)

Explanation:
1. With \( \omega=\frac{qB_0}{m} \): \( \dot v_x=\omega v_y \) and \( \dot v_y=\frac{qE_0}{m}-\omega v_x \).
2. The solution (cycloid) starting from rest is \( v_y=\frac{E_0}{B_0}\sin\omega t \), \( v_x=\frac{E_0}{B_0}(1-\cos\omega t) \).
3. Integrating \( v_y \): \( y=\frac{E_0}{B_0\omega}(1-\cos\omega t)=\frac{E_0m}{qB_0^2}(1-\cos\omega t) \).

Teacher's Note:
1. Draw the directions of the fields and forces first; then use magnitudes for the numbers.

Answer: (c) \( \frac15q\omega R^2 \)

Explanation:
1. A shell of radius r and thickness dr carries charge \( dq=\frac{3q}{R^3}r^2dr \) and has moment \( \frac13dq\,\omega r^2 \).
2. Integrating: \( \mu=\frac{q\omega}{R^3}\int_0^Rr^4dr=\frac15q\omega R^2 \).

Teacher's Note:
1. Draw the directions of the fields and forces first; then use magnitudes for the numbers.

Answer: (a) \( 1.8\times10^{-4} \) T

Explanation:
1. For a short dipole \( B=\frac{\mu_0}{4\pi}\frac{M}{r^3}\sqrt{1+3\cos^2\theta} \).
2. With \( \cos^230^\circ=\frac34 \): \( \sqrt{1+\frac94}=1.80 \).
3. So \( B=10^{-7}\times\frac{1}{10^{-3}}\times1.80=1.8\times10^{-4} \) T.

Teacher's Note:
1. Draw the directions of the fields and forces first; then use magnitudes for the numbers.

Answer: (c) \( \frac{4\times10^{-13}}{27}\left(\hat k-2\hat j\right) \) T

Explanation:
1. Here \( \vec r=2\hat i+\hat j+2\hat k \) and \( r=3 \) m, so \( r^3=27 \).
2. Cross product: \( \vec v\times\vec r=2\hat i\times(2\hat i+\hat j+2\hat k)=2(\hat k-2\hat j) \).
3. So \( \vec B=10^{-7}\times2\times10^{-6}\times\frac{2(\hat k-2\hat j)}{27}=\frac{4\times10^{-13}}{27}(\hat k-2\hat j) \) T.

Teacher's Note:
1. Draw the directions of the fields and forces first; then use magnitudes for the numbers.

Answer: (a) \( 2\times10^{-5} \) T

Explanation:
1. Each wire gives \( \frac{\mu_0i}{2\pi r}=\frac{2\times10^{-7}\times10}{0.1}=2\times10^{-5} \) T, and the two fields are perpendicular to the lines joining the point to each wire.
2. The point and the wires form an equilateral triangle, so the fields are at 120 degrees to each other; for equal magnitudes the resultant equals each one: \( 2\times10^{-5} \) T.

Teacher's Note:
1. Draw the directions of the fields and forces first; then use magnitudes for the numbers.

Answer: (c) \( \frac{2\sqrt2\,\mu_0i}{\pi a} \)

Explanation:
1. Each side is at perpendicular distance \( \frac a2 \) and subtends \( 45^\circ \) at each end: \( \frac{\mu_0i}{4\pi(a/2)}\cdot2\sin45^\circ=\frac{\mu_0i}{\sqrt2\pi a} \).
2. Four sides add: \( B=\frac{4\mu_0i}{\sqrt2\pi a}=\frac{2\sqrt2\mu_0i}{\pi a} \).

Teacher's Note:
1. Draw the directions of the fields and forces first; then use magnitudes for the numbers.

Answer: (a) \( 8.9\times10^{-4} \) T

Explanation:
1. After rotation the two fields are perpendicular, each of magnitude \( 6.28\times10^{-4} \) T.
2. Resultant \( \sqrt2\times6.28\times10^{-4}=8.9\times10^{-4} \) T.

Teacher's Note:
1. Draw the directions of the fields and forces first; then use magnitudes for the numbers.

Answer: (c) \( \frac{2\mu_0i}{5\pi r} \)

Explanation:
1. Outside a long cylindrical current the field is that of a line current on the axis: \( B=\frac{\mu_0i}{2\pi d} \) with \( d=r+\frac r4=\frac{5r}4 \).
2. So \( B=\frac{\mu_0i}{2\pi}\cdot\frac{4}{5r}=\frac{2\mu_0i}{5\pi r} \). (Inside the tube, at r/4 from the surface, the field is zero.)

Teacher's Note:
1. Draw the directions of the fields and forces first; then use magnitudes for the numbers.

Answer: (a) 1.0 V

Explanation:
1. Turn density \( n=\frac{400}{0.2}=2000 \) per metre, so \( i=\frac{B}{\mu_0n}=\frac{10^{-2}}{4\pi\times10^{-7}\times2000}=3.98 \) A.
2. Wire length \( 400\times2\pi\times0.01=25.1 \) m, resistance \( 0.251 \) ohm.
3. emf \( =iR=3.98\times0.251=1.0 \) V.

Teacher's Note:
1. Draw the directions of the fields and forces first; then use magnitudes for the numbers.

Answer: (c) \( (-75\hat i+100\hat j) \) m/s

Explanation:
1. With \( \vec v=v_x\hat i+v_y\hat j \): \( \vec v\times\vec B=B(v_y\hat i-v_x\hat j) \) with \( B=8\times10^{-3} \) T.
2. Force/charge \( =(0.8\hat i+0.6\hat j) \) so \( 8\times10^{-3}v_y=0.8 \) gives \( v_y=100 \) and \( -8\times10^{-3}v_x=0.6 \) gives \( v_x=-75 \) m/s.

Teacher's Note:
1. Draw the directions of the fields and forces first; then use magnitudes for the numbers.

Answer: (a) 50 m/s

Explanation:
1. Uniform velocity means the magnetic force balances the weight: \( qvB=mg \).
2. So \( v=\frac{mg}{qB}=\frac{10^{-5}\times10}{10^{-5}\times0.2}=50 \) m/s.

Teacher's Note:
1. Draw the directions of the fields and forces first; then use magnitudes for the numbers.

Answer: (c) \( 2.0\times10^5 \) m/s and \( 1.27\times10^5 \) m/s

Explanation:
1. Perpendicular: \( v_\perp=\frac{qBr}{m}=\frac{1.6\times10^{-19}\times0.04\times0.05}{1.6\times10^{-27}}=2.0\times10^5 \) m/s.
2. Period \( T=\frac{2\pi m}{qB}=1.57\times10^{-6} \) s, so \( v_\parallel=\frac{0.20}{1.57\times10^{-6}}=1.27\times10^5 \) m/s.

Teacher's Note:
1. Draw the directions of the fields and forces first; then use magnitudes for the numbers.