JEE Physics Electromagnetic Induction MCQs Set 12

Answer: (a) 0.058 T

Explanation:
1. \( \phi=BA\sin60^\circ \) because the angle is measured from the plane, so the normal makes 30° with the field.
2. \( B=\dfrac{10^{-3}}{0.02\times0.866}\approx0.058 \) T.

Teacher's Note:
1. The flux uses the angle between the field and the normal to the surface.
2. Angle with the plane is the complement.

Answer: (a) \( 2\pi(R_0+t)B \), with the induced current anticlockwise

Explanation:
1. \( \varepsilon=B\dfrac{dA}{dt}=B\cdot2\pi R\dfrac{dR}{dt}=2\pi(R_0+t)B \).
2. The inward flux increases, so by Lenz's law the induced current produces an outward field: anticlockwise.

Teacher's Note:
1. Differentiate the area with respect to time.
2. Lenz's law: the induced current opposes the increase of flux.

Answer: (c) 0.2 A

Explanation:
1. \( \varepsilon=-\dfrac{d\phi}{dt}=-(12t-5) \); at \( t=0.25 \): \( \varepsilon=2 \) V in magnitude.
2. Current \( =2/10=0.2 \) A.

Teacher's Note:
1. Differentiate the flux, then divide by the resistance.
2. Take the magnitude unless direction is asked.

Answer: (c) \( \dfrac{\mu_0er^2}{4R}\alpha \)

Explanation:
1. Equivalent current \( I=\dfrac{e\omega}{2\pi} \), so \( \dfrac{dI}{dt}=\dfrac{e\alpha}{2\pi} \).
2. Field at the centre \( B=\dfrac{\mu_0I}{2R} \) acts over the small loop: \( \phi=\dfrac{\mu_0I}{2R}\pi r^2 \).
3. Emf \( =\dfrac{\mu_0\pi r^2}{2R}\cdot\dfrac{e\alpha}{2\pi}=\dfrac{\mu_0er^2\alpha}{4R} \).

Teacher's Note:
1. A changing speed changes the equivalent current and hence the field.
2. Treat the field as uniform over the tiny loop.

Answer: (b) End Y becomes positively charged

Explanation:
1. Force on positive charges: \( q\vec v\times\vec B \) with \( \vec v=\hat x \), \( \vec B=-\hat z \) gives \( +\hat y \), upwards.
2. So positive charge accumulates at the top end Y until the electric force balances the magnetic force.
3. No current flows in an isolated rod, so there is no heating.

Teacher's Note:
1. Motional emf appears across the ends of a moving conductor.
2. Check direction with the cross product.

Answer: (d) remain the same

Explanation:
1. Motional emf depends on the effective length, the straight distance between the ends of the wire, \( \varepsilon=Bv\ell_{eff} \).
2. The end points are unchanged, so the emf and the current are unchanged.

Teacher's Note:
1. Only the perpendicular separation of the ends matters.
2. The shape of the moving wire is irrelevant.

Answer: (d) zero

Explanation:
1. The flux through the loop stays constant (\( B\ell^2 \)).
2. The emfs in the two opposite sides moving in the field are equal and opposite around the loop, so the net emf and the current are zero.

Teacher's Note:
1. Look at the net change of flux, not at individual sides.
2. A loop fully inside a uniform field has no net emf.

Answer: (b) \( \dfrac{\mu_0bI}{2\pi\tau}\ln\dfrac{d+a}{d} \)

Explanation:
1. Flux \( \phi=\int_d^{d+a}\dfrac{\mu_0I}{2\pi x}b\,dx=\dfrac{\mu_0Ib}{2\pi}\ln\dfrac{d+a}{d} \).
2. Emf \( =\left|\dfrac{d\phi}{dt}\right|=\dfrac{\mu_0b}{2\pi}\ln\dfrac{d+a}{d}\cdot\dfrac{I}{\tau} \) since \( dI/dt=-I/\tau \).

Teacher's Note:
1. Integrate the field of the wire over the loop area.
2. Differentiate the exponential current.

Answer: (a) \( \dfrac{V_0B_0d^2}{a} \)

Explanation:
1. The two sides parallel to y are at \( x \) and \( x+d \), with fields differing by \( B_0d/a \).
2. Motional emfs \( V_0d\,B(x+d) \) and \( V_0d\,B(x) \) oppose, so \( \varepsilon=V_0d\cdot\dfrac{B_0d}{a}=\dfrac{V_0B_0d^2}{a} \).

Teacher's Note:
1. The emfs of the two sides subtract.
2. The sides parallel to x have velocity along their length, so they give no emf.

Answer: (b) \( v_x\ell_2-v_y\ell_1 \)

Explanation:
1. \( \vec v\times\vec B=B(v_y\hat i-v_x\hat j) \) for \( \vec B=B\hat k \).
2. Emf along AB (x direction): \( Bv_y\ell_1 \). Emf along BC (y direction): \( -Bv_x\ell_2 \).
3. The total magnitude is \( B(v_x\ell_2-v_y\ell_1) \).

Teacher's Note:
1. Resolve the velocity along and perpendicular to each segment.
2. Only the component of velocity perpendicular to the segment contributes.

Answer: (b) zero

Explanation:
1. Each half of the rod has emf \( \tfrac12B\omega(\ell/2)^2 \) with the same sense measured from the centre.
2. Both ends are at the same potential relative to the centre, so the difference between the ends is zero.

Teacher's Note:
1. The two halves cancel by symmetry.
2. Rotation about the end would give \( \tfrac12B\omega\ell^2 \).

Answer: (d) in the anticlockwise direction

Explanation:
1. As the north pole approaches, the flux through the coil towards the observer increases.
2. The induced current opposes this by creating a field away from the observer: the face towards the magnet becomes a north pole.
3. A north-pole face corresponds to an anticlockwise current when seen from that side.

Teacher's Note:
1. Opposing the approach means the nearer face acts as the same pole.
2. Anticlockwise current gives a north pole on the viewing side.

Answer: (b) 75 mH

Explanation:
1. \( L\propto N^2 \).
2. \( L'=108\times(500/600)^2=75 \) mH.

Teacher's Note:
1. Inductance varies as the square of the number of turns.
2. Same geometry otherwise.

Answer: (d) \( \dfrac{a^2T^3}{3R} \)

Explanation:
1. Emf \( \varepsilon=\left|\dfrac{d\phi}{dt}\right|=a(T-2t) \).
2. Heat \( H=\int_0^T\dfrac{\varepsilon^2}{R}dt=\dfrac{a^2}{R}\int_0^T(T-2t)^2dt=\dfrac{a^2T^3}{3R} \).

Teacher's Note:
1. Heat is the integral of \( \varepsilon^2/R \) over time.
2. Integrate between 0 and T.

Answer: (a) \( e=E-iR \), a straight line of negative slope

Explanation:
1. Kirchhoff's loop rule: \( E-iR-e=0 \).
2. Hence \( e=E-iR \), a straight line with intercept \( E \) at \( i=0 \) and slope \( -R \).

Teacher's Note:
1. Use the loop rule at any instant.
2. At t = 0 the whole emf appears across the inductor.

Answer: (b) 13.5% of the maximum

Explanation:
1. \( i=i_0e^{-Rt/L}=i_0e^{-2} \).
2. \( e^{-2}=0.135 \), so 13.5%.

Teacher's Note:
1. The current falls to 36.8% in one time constant.
2. After two time constants: \( e^{-2} \).

Answer: (c) the emf induced in the inductor

Explanation:
1. Just after closing, the inductor opposes the sudden change: current is zero.
2. Energy \( \tfrac12Li^2 \) is zero, power \( Ei \) is zero.
3. The induced emf equals the battery emf, so it is not zero.

Teacher's Note:
1. At t = 0 the inductor behaves like an open circuit.
2. The full battery emf appears across it.

Answer: (c) \( \dfrac{|F_1-F_2|\,mL}{eM} \)

Explanation:
1. The rod accelerates with \( a=\dfrac{|F_1-F_2|}{M} \).
2. The free electrons must be accelerated by an electric field: \( eE=ma_e=m a \), so \( E=\dfrac{ma}{e} \).
3. Potential difference \( =EL=\dfrac{|F_1-F_2|\,mL}{eM} \).

Teacher's Note:
1. Electrons move with the rod only if a field accelerates them.
2. V = E times length.

Answer: (b) \( \dfrac{Bv}{r} \)

Explanation:
1. At time \( t \) the side is \( s=\ell+2vt \), so the area \( s^2 \) grows at \( \dfrac{dA}{dt}=2s\cdot2v=4sv \).
2. Emf \( \varepsilon=4Bvs \); resistance \( R=4sr \).
3. Current \( i=\dfrac{4Bvs}{4sr}=\dfrac{Bv}{r} \).

Teacher's Note:
1. Both pairs of conductors move, so each side grows at 2v.
2. The side length cancels.

Answer: (a) it moves downward with decreasing acceleration and finally acquires a constant velocity

Explanation:
1. As the speed grows, the emf \( B\ell v \) and the current grow, giving an upward magnetic force \( B^2\ell^2v/R \).
2. The net force \( mg-B^2\ell^2v/R \) falls, so the acceleration decreases.
3. At terminal velocity \( v_T=\dfrac{mgR}{B^2\ell^2} \) the force balances and the acceleration is zero.

Teacher's Note:
1. Magnetic drag grows with speed.
2. Terminal velocity is reached when drag equals weight.

Answer: (c) \( \dfrac{eR}{2m}\dfrac{dB}{dt} \)

Explanation:
1. Induced field at radius \( R \): \( E\cdot2\pi R=\pi R^2\dfrac{dB}{dt}\Rightarrow E=\dfrac R2\dfrac{dB}{dt} \).
2. Acceleration \( a=\dfrac{eE}{m}=\dfrac{eR}{2m}\dfrac{dB}{dt} \).

Teacher's Note:
1. A changing magnetic field produces a circulating electric field.
2. Use Faraday's law on a circular path.