JEE Physics Moving Charges and Magnetism MCQs Set 16

Answer: (d) \( \frac{\mu_0Ir}{2\pi a^2} \)

Explanation:
1. Take an Amperian loop of radius r < a: \( B\cdot2\pi r=\mu_0 I\frac{r^2}{a^2} \).
2. So \( B=\frac{\mu_0Ir}{2\pi a^2} \).
3. The field rises linearly from zero at the axis.

Teacher's Note:
1. At r = a this gives \( \frac{\mu_0I}{2\pi a} \), matching the outside result.
2. Outside, B falls as 1/r.

Answer: (a) \( \frac{\mu_0I}{\pi^2R} \)

Explanation:
1. Divide the semicircle into thin wires; the element at angle \( \theta \) carries \( dI=\frac{I}{\pi}d\theta \).
2. Its field at the axis has magnitude \( \frac{\mu_0dI}{2\pi R} \); by symmetry the components along the diameter cancel and those perpendicular add: \( dB=\frac{\mu_0I}{2\pi^2R}\sin\theta\,d\theta \).
3. Integrating from 0 to \( \pi \): \( B=\frac{\mu_0I}{2\pi^2R}\times2=\frac{\mu_0I}{\pi^2R} \).

Teacher's Note:
1. The integral of sin from 0 to pi is 2.
2. Use symmetry before integrating.

Answer: (c) B is proportional to \( 1/R \) (a falling hyperbola)

Explanation:
1. Charge density: \( \sigma=\frac{Q}{\pi R^2} \).
2. Field at the centre of a rotating charged disc: \( B=\int_0^R\frac{\mu_0\,dI}{2x}=\frac{\mu_0\sigma\omega R}{2} \).
3. Substituting \( \sigma \): \( B=\frac{\mu_0Q\omega}{2\pi R} \), so \( B\propto\frac1R \).

Teacher's Note:
1. Because the total charge is fixed, a bigger disc has a smaller surface density.
2. Do not use B proportional to R, which holds at fixed density.

Answer: (d) 2.97 W

Explanation:
1. Force on the conductor (length 3 m): \( F=iLB(x)=10\times3\times3\times10^{-4}e^{-0.2x}=9\times10^{-3}e^{-0.2x} \) N.
2. Work in moving it through 2 m: \( W=\int_0^2F\,dx=9\times10^{-3}\cdot\frac{1-e^{-0.4}}{0.2}=1.48\times10^{-2} \) J.
3. Power: \( P=\frac{W}{t}=\frac{1.48\times10^{-2}}{5\times10^{-3}}=2.97 \) W.

Teacher's Note:
1. At constant speed the applied force balances the magnetic force.
2. Remember to integrate because the field varies with x.

Answer: (b) \( \frac{\pi^2}{8\sqrt2} \)

Explanation:
1. Circle: \( l=2\pi R \), so \( B_A=\frac{\mu_0I}{2R}=\frac{\mu_0I\pi}{l} \).
2. Square: \( a=\frac l4 \); each side at perpendicular distance \( a/2 \) gives \( \frac{\mu_0I}{4\pi(a/2)}\cdot2\sin45^\circ=\frac{\mu_0I\sqrt2}{2\pi a} \); four sides: \( B_B=\frac{2\sqrt2\mu_0I}{\pi a}=\frac{8\sqrt2\mu_0I}{\pi l} \).
3. Ratio: \( \frac{B_A}{B_B}=\frac{\pi}{8\sqrt2/\pi}=\frac{\pi^2}{8\sqrt2} \).

Teacher's Note:
1. Equal lengths of wire means equal perimeters.
2. A circle gives a larger field than a square for the same wire length, but the ratio here is just about 0.87.

Answer: (c) \( r_e<r_p=r_\alpha \)

Explanation:
1. For kinetic energy K: \( r=\frac{mv}{qB}=\frac{\sqrt{2mK}}{qB} \).
2. Compare \( \frac{\sqrt m}{q} \): electron is very small (tiny m); proton \( \frac{\sqrt m}{e} \); alpha \( \frac{\sqrt{4m}}{2e}=\frac{\sqrt m}{e} \).
3. So the proton and alpha have equal radii and the electron has a smaller one.

Teacher's Note:
1. The alpha has four times the proton mass and twice its charge, which cancels in the ratio.
2. Mass of electron is about 1/1836 of the proton.

Answer: (d) \( \sqrt2 \)

Explanation:
1. Moment \( m=I\pi R^2 \): doubling m at fixed I multiplies \( R^2 \) by 2, so \( R_2=\sqrt2R_1 \).
2. Field at the centre: \( B=\frac{\mu_0I}{2R}\propto\frac1R \).
3. Therefore \( \frac{B_1}{B_2}=\frac{R_2}{R_1}=\sqrt2 \).

Teacher's Note:
1. Keep track of which quantity is held constant.
2. The field falls as the loop grows.

Answer: (b) 12.87 cm

Explanation:
1. Radius: \( r=\frac{\sqrt{2mK}}{eB}=\frac{5.40\times10^{-24}}{2.4\times10^{-22}}=2.25 \) cm.
2. Deflection angle on leaving the 2 cm region: \( \sin\theta=\frac{2}{2.25}=0.89 \), so \( \theta=62.8^\circ \). Sideways shift inside the field: \( r(1-\cos\theta)=1.22 \) cm.
3. Straight path over the remaining 6 cm adds \( 6\tan\theta=11.7 \) cm. Total \( d\approx1.22+11.7=12.9 \) cm, so the option closest to this is 12.87 cm.

Teacher's Note:
1. The 1.22 cm and 2.25 cm options come from stopping the calculation too early.
2. Compute the exit angle from the geometry of the arc.

Answer: (c) \( \frac{\sqrt3\,mv_0}{qE_0} \)

Explanation:
1. The magnetic field is along x, so it only rotates the velocity components in the y-z plane and does not change that component's magnitude: \( v_y^2+v_z^2=v_0^2 \).
2. The electric field along x gives \( v_x=\frac{qE_0}{m}t \).
3. Speed squared: \( v^2=v_0^2+\left(\frac{qE_0t}{m}\right)^2=4v_0^2 \), so \( t=\frac{\sqrt3mv_0}{qE_0} \).

Teacher's Note:
1. B along x cannot change motion along x.
2. Doubling the speed means quadrupling the squared speed.

Answer: (a) \( \frac23 \)

Explanation:
1. At \( r=a/3 \) (inside): \( B=\frac{\mu_0Ir}{2\pi a^2}=\frac{\mu_0I}{6\pi a} \).
2. At \( r=2a \) (outside): \( B=\frac{\mu_0I}{2\pi(2a)}=\frac{\mu_0I}{4\pi a} \).
3. Ratio: \( \frac{1/6}{1/4}=\frac23 \).

Teacher's Note:
1. Use the inside formula for r < a and the outside one for r > a.
2. Check continuity at r = a.

Answer: (b) \( \frac{e\mu_0nIR}{2m} \)

Explanation:
1. Field inside: \( B=\mu_0nI \). The electron moves on a circle of radius \( r=\frac{mv}{eB} \) starting from the axis.
2. The circle passes through the axis and its farthest point from the axis is at distance \( 2r \).
3. To avoid hitting the wall: \( 2r\le R \), so \( v\le\frac{eBR}{2m}=\frac{e\mu_0nIR}{2m} \).

Teacher's Note:
1. The diameter of the circle is the farthest reach from the starting point.
2. Strike the wall when 2r = R.

Answer: (c) \( T=\sqrt{\frac{2\pi m}{IB}} \)

Explanation:
1. Restoring torque for a small angle: \( \tau=-\mu B\theta=-I\pi a^2B\,\theta \).
2. Moment of inertia of the ring about a diameter: \( I_d=\frac{ma^2}{2} \).
3. So \( \omega^2=\frac{I\pi a^2B}{ma^2/2}=\frac{2\pi IB}{m} \) and \( T=\frac{2\pi}{\omega}=\sqrt{\frac{2\pi m}{IB}} \).

Teacher's Note:
1. Use the moment of inertia about a diameter, not about the axis.
2. Same method as a magnetic needle in a field.

Answer: (a) \( \frac{\mu_0NI}{2(b-a)}\ln\frac{b}{a} \)

Explanation:
1. Number of turns in a thin annulus of radius r and width dr: \( dn=\frac{N}{b-a}dr \).
2. Field of those turns at the centre: \( dB=\frac{\mu_0\,dn\,I}{2r}=\frac{\mu_0NI}{2(b-a)}\frac{dr}{r} \).
3. Integrate from a to b: \( B=\frac{\mu_0NI}{2(b-a)}\ln\frac ba \).

Teacher's Note:
1. Treat the spiral as many concentric rings.
2. If a approaches b the result approaches the field of N loops: \( \frac{\mu_0NI}{2b} \).

Answer: (b) move in a straight line with constant velocity

Explanation:
1. Electric force on the electron: \( -e\vec E=+eE_0\hat x \).
2. Magnetic force: \( -e(\vec v\times\vec B)=-e\frac{E_0}{B_0}B_0(\hat y\times\hat z)=-eE_0\hat x \).
3. The two forces cancel, so the net force is zero and the electron moves in a straight line with constant velocity.

Teacher's Note:
1. This is the velocity selector condition \( v=E/B \).
2. The speed E/B passes undeflected for any charge.

Answer: (d) \( 2qV \)

Explanation:
1. The particle crosses the gap twice in each revolution, and the revolution period equals the period of the voltage.
2. Each crossing adds \( qV \) (voltage at its peak).
3. So in one time period the gain is \( 2qV \).

Teacher's Note:
1. The question counts a full period, not a half period.
2. Use peak voltage for the gain per crossing.

Answer: (a) \( 2\pi \) T

Explanation:
1. Matching condition: \( qB=2\pi mf_{osc} \), so \( B=\frac{2\pi f_{osc}}{q/m} \).
2. Substituting: \( B=\frac{2\pi\times10^6}{10^6}=2\pi \) T.

Teacher's Note:
1. The frequency of revolution is independent of speed.
2. Units: Hz divided by C/kg gives tesla.

Answer: (b) changing the applied magnetic field (to satisfy the rynchronisation condition for the fixed oscillator frequency)

Explanation:
1. For fixed oscillator frequency, rynchronisation needs \( qB=2\pi mf_{osc} \).
2. Since q and m are fixed for a given particle, B must be varied until this holds.
3. The voltage amplitude only fixes the energy gained per crossing, not the rynchronisation.

Teacher's Note:
1. In the passage the oscillator frequency is fixed, so B is what is adjusted.
2. The voltage amplitude affects how many turns are needed, not the tuning.

Answer: (d) \( \frac{3d}{4} \)

Explanation:
1. Radius: \( r=\frac{mv}{qB}=\frac{v_m}{4}\frac{m}{qB}=\frac d8 \).
2. Positions at turning angle \( \phi \): the horizontal separation is \( d-2r\sin\phi \) with equal heights.
3. Minimum at \( \phi=90^\circ \): \( d-2r=d-\frac d4=\frac{3d}{4} \).

Teacher's Note:
1. The particles mirror each other, so the vertical separation is always zero.
2. Maximum separation is at \( \phi=270^\circ \).

Answer: (a) \( \frac{5d}{4} \)

Explanation:
1. Radius \( r=\frac d8 \) as before; separation along the line joining them is \( d-2r\sin\phi \).
2. The sine reaches \( -1 \) at \( \phi=270^\circ \) (the particles have turned back and moved apart).
3. So the maximum separation is \( d+2r=d+\frac d4=\frac{5d}{4} \).

Teacher's Note:
1. Over the full circle, the separation oscillates between d - 2r and d + 2r.
2. They never collide since r < d/2.

Answer: (c) a straight line

Explanation:
1. At collision the two velocities have equal components along the bisector (same sense, upward) and equal and opposite components along the line joining them.
2. The combined body has charge \( +q-q=0 \), so it feels no magnetic force.
3. With no net force (momentum conserved), it moves in a straight line at constant velocity.

Teacher's Note:
1. A neutral object is unaffected by the magnetic field.
2. The horizontal momenta cancel, leaving motion along the bisector.

Answer: (d) 3.6

Explanation:
1. Frequency \( f=\frac{\omega}{2\pi}=5\times10^3 \) Hz, so the equivalent current is \( i=qf=5000 \) A on a circle of radius 0.6 m.
2. On the axis: \( B=\frac{\mu_0iR^2}{2(R^2+x^2)^{3/2}}=\frac{4\pi\times10^{-7}\times5000\times0.36}{2\times1} \).
3. So \( B=3.6\pi\times10^{-4} \) T and N = 3.6.

Teacher's Note:
1. Here \( R^2+x^2=0.36+0.64=1 \), which simplifies the arithmetic.
2. A rotating point charge acts as a current loop of strength qf.

Answer: (a) 2

Explanation:
1. Since \( \ell\gg d \), the two long sides behave like infinite wires, each at distance \( d/2 \) from the centre and carrying current in opposite senses on opposite sides, so their fields add.
2. Each gives \( \frac{\mu_0i}{2\pi(d/2)}=\frac{\mu_0i}{\pi d} \).
3. Total \( \frac{2\mu_0i}{\pi d} \), so N = 2 (the short sides contribute negligibly).

Teacher's Note:
1. Currents in opposite directions on opposite sides produce fields in the same direction at the centre.
2. The short sides contribute a term of order d/l.

Answer: (c) 3

Explanation:
1. Enclosed current: \( i=\int_0^{r_1}br\cdot2\pi r\,dr=\frac{2\pi br_1^3}{3} \).
2. Ampere's law: \( B\,2\pi r_1=\mu_0i \).
3. So \( B=\frac{\mu_0br_1^2}{3} \) and N = 3.

Teacher's Note:
1. For uniform J the enclosed current would be proportional to r squared; here it goes as r cubed.
2. Always integrate \( J\,dA \) over rings.

Answer: (d) 16

Explanation:
1. Initial charge: \( Q=CV=50\times10^{-6}\times20=10^{-3} \) C. A fall of 10% in voltage means \( \Delta Q=10^{-4} \) C.
2. Average current: \( i=\frac{\Delta Q}{t}=\frac{10^{-4}}{2}=5\times10^{-5} \) A.
3. Field: \( B=\mu_0ni=4\pi\times10^{-7}\times8000\times5\times10^{-5}=16\pi\times10^{-8} \) T, so N = 16.

Teacher's Note:
1. Voltage across a capacitor is proportional to its charge.
2. Use the average current over the 2 s.

Answer: (b) 2

Explanation:
1. At rest only the electric field acts: \( eE=ma_0 \) (east).
2. With velocity \( v_0 \) the magnetic force must supply the extra acceleration \( 2a_0 \) towards east: \( ev_0B_\perp=2ma_0 \).
3. The smallest field magnitude is when B is perpendicular to the velocity: \( B=\frac{2ma_0}{ev_0} \), N = 2.

Teacher's Note:
1. A component of B along v would add nothing to the force, so it would only increase |B|.
2. Subtract the electric contribution first.