JEE Physics Electromagnetic Induction MCQs Set 13

Answer: (c) \( \dfrac{2\omega BA}{\pi} \)

Explanation:
1. Flux changes from \( BA \) to 0.
2. Time taken \( t=\dfrac{\pi/2}{\omega} \).
3. Average emf \( =\dfrac{BA}{\pi/(2\omega)}=\dfrac{2\omega BA}{\pi} \).

Teacher's Note:
1. Average emf is the change of flux divided by the time.
2. A quarter turn takes \( \pi/2\omega \).

Answer: (b) \( \dfrac{\mu_0\pi r^2}{2R} \)

Explanation:
1. Field at the centre of the large coil per unit current: \( \dfrac{\mu_0}{2R} \).
2. Flux through the small coil: \( \dfrac{\mu_0}{2R}\pi r^2 \); so \( M=\dfrac{\mu_0\pi r^2}{2R} \).

Teacher's Note:
1. Treat the field over the small coil as uniform.
2. M is flux per unit current.

Answer: (a) inertia

Explanation:
1. Self-inductance opposes any change of current, just as mass resists change of velocity.
2. Hence it plays the role of inertia.

Teacher's Note:
1. Compare \( \tfrac12Li^2 \) with \( \tfrac12mv^2 \).
2. Current is the analogue of velocity.

Answer: (b) 4 mH

Explanation:
1. For perfect coupling \( M=\sqrt{L_1L_2}=\sqrt{2\times8} \) mH = 4 mH.

Teacher's Note:
1. Perfect coupling means coupling coefficient k = 1.
2. Geometric mean.

Answer: (d) 0.02 H

Explanation:
1. \( L=L_1+L_2-2M=0.01+0.03-0.02=0.02 \) H.

Teacher's Note:
1. Opposing fluxes subtract twice the mutual inductance.
2. Aiding fluxes add it.

Answer: (a) \( \dfrac{1}{220} \) A

Explanation:
1. Emf \( \varepsilon=B\ell v=0.1\times0.1\times1=0.01 \) V.
2. The 2 Ω and 3 Ω resistors are in parallel: \( \dfrac{2\times3}{5}=1.2 \) Ω; in series with the connector 1 Ω: total 2.2 Ω.
3. Current \( =\dfrac{0.01}{2.2}=\dfrac{1}{220} \) A.

Teacher's Note:
1. The connector acts as the source with its own internal resistance.
2. Combine the parallel resistors first.

Answer: (b) \( \dfrac{LE^2}{2R_1^2} \)

Explanation:
1. In steady state the current is \( I=E/R_1 \).
2. The stored energy \( \tfrac12LI^2=\dfrac{LE^2}{2R_1^2} \) is dissipated entirely in the resistor in the closed discharge loop, which is \( R_2 \).

Teacher's Note:
1. All the stored magnetic energy turns into heat.
2. The current just before the switch is E/R₁.

Answer: (b) \( \left(\dfrac{e^2}{e^2+1}\right)^2 \)

Explanation:
1. Bulb resistance \( R=\dfrac{200^2}{100}=400 \) Ω; \( \tau=L/R=0.5 \) s.
2. Current \( i=i_0(1-e^{-2t}) \); power \( \propto i^2 \).
3. Ratio \( \left(\dfrac{1-e^{-2}}{1-e^{-4}}\right)^2=\left(\dfrac{1}{1+e^{-2}}\right)^2=\left(\dfrac{e^2}{e^2+1}\right)^2 \).

Teacher's Note:
1. Find the time constant from the bulb's resistance.
2. Use growth of current in an LR circuit.

Answer: (b) \( 3.125\times10^{-3} \) J

Explanation:
1. Speed \( v=0.5/2=0.25 \) m/s; emf \( B\ell v=0.125 \) V.
2. Power \( \dfrac{\varepsilon^2}{R}=1.5625\times10^{-3} \) W.
3. Work \( =P\,t=3.125\times10^{-3} \) J.

Teacher's Note:
1. Work done equals the electrical energy dissipated.
2. Use ε²/R for power.

Answer: (c) \( Bi\ell=mg\tan\theta \)

Explanation:
1. The magnetic force \( Bi\ell \) is horizontal (field vertical, wire horizontal).
2. Components along the incline: \( Bi\ell\cos\theta=mg\sin\theta \).
3. Hence \( Bi\ell=mg\tan\theta \).

Teacher's Note:
1. Resolve forces along the incline, since the rails are smooth.
2. A vertical field gives a horizontal force.

Answer: (d) (2) and (4) only

Explanation:
1. Just after closing the capacitor is uncharged, and the inductor carries no current, so (1) and (3) are false.
2. After a long time the capacitor is fully charged to \( E \) so \( q=EC \), and the inductor behaves as a plain wire with \( i=E/R \); (2) and (4) are true.

Teacher's Note:
1. Capacitor: initially a short, finally an open circuit.
2. Inductor: initially open, finally a short.

Answer: (a) (1), (3) and (4) only

Explanation:
1. Power \( P=Li\dfrac{di}{dt} \) is equal, so \( L_1I_1=L_2I_2 \) and \( I_1/I_2=L_2/L_1=1/4 \): (1) true, (2) false.
2. Voltage \( V=L\dfrac{di}{dt} \): \( V_2/V_1=L_2/L_1=1/4 \): (4) true.
3. Energy \( W=\tfrac12LI^2 \): \( \dfrac{W_2}{W_1}=\dfrac{2\times16}{8\times1}=4 \): (3) true.

Teacher's Note:
1. Same di/dt for both coils.
2. Use P = L i di/dt, V = L di/dt and W = ½Li².

Answer: (b) 40 rad/s

Explanation:
1. Induced electric field at the ring: \( E=\dfrac R2\dfrac{dB}{dt}=\dfrac22\times0.2=0.2 \) V/m.
2. Tangential force on the total charge \( F=qE=0.4 \) N, so \( a=F/m=0.4/0.05=8 \) m/s².
3. Speed \( v=at=80 \) m/s, so \( \omega=v/R=80/2=40 \) rad/s.

Teacher's Note:
1. Changing flux drives a tangential electric field around the ring.
2. Angular speed is v/R.

Answer: (d) \( 1\times10^{-5} \) V

Explanation:
1. Vertical component \( B_V=B_H\tan\delta=3\times10^{-4}\times\tfrac43=4\times10^{-4} \) T.
2. Only the vertical component gives an emf along a north-south rod moving east: \( \varepsilon=B_Vv\ell=4\times10^{-4}\times0.1\times0.25=1\times10^{-5} \) V.

Teacher's Note:
1. Use the vertical component of the earth's field.
2. Check that v, B and the rod are mutually perpendicular.

Answer: (b) \( \dfrac{mgR}{B^2L^2} \)

Explanation:
1. At terminal velocity the tension \( mg \) balances the magnetic force \( BiL=\dfrac{B^2L^2v}{R} \).
2. So \( v_T=\dfrac{mgR}{B^2L^2} \).

Teacher's Note:
1. The rod is massless, so the tension equals the magnetic force at all times.
2. Terminal velocity: zero acceleration.

Answer: (d) \( g/2 \)

Explanation:
1. The rod is massless, so the string tension equals the magnetic force \( \dfrac{B^2L^2v}{R} \), which at \( v_T/2 \) equals \( mg/2 \).
2. For the hanging mass: \( mg-\dfrac{mg}2=ma \), so \( a=g/2 \).

Teacher's Note:
1. Magnetic force is proportional to speed.
2. At half terminal speed it is half the weight.

Answer: (b) \( \dfrac{mgR}{B^2\ell^2} \)

Explanation:
1. At terminal speed \( mg=B i\ell=\dfrac{B^2\ell^2v}{R} \).
2. So \( v=\dfrac{mgR}{B^2\ell^2} \).

Teacher's Note:
1. Weight balances the magnetic force.
2. Current \( i=B\ell v/R \).

Answer: (d) \( \dfrac{e^2}{e^2-1} \)

Explanation:
1. Time constant \( \tau=L/R=0.5 \) s; \( i=i_0(1-e^{-t/\tau}) \).
2. At 1 s: \( i=i_0(1-e^{-2}) \).
3. Ratio \( \dfrac{i_\infty}{i_1}=\dfrac{1}{1-e^{-2}}=\dfrac{e^2}{e^2-1}\approx1.16 \).

Teacher's Note:
1. The final current is E/R.
2. Use the growth law for an LR circuit.

Answer: (c) 100 m

Explanation:
1. For a wire of length \( \ell_w \): \( N\cdot2\pi r=\ell_w \), \( L=\dfrac{\mu_0N^2\pi r^2}{l_0}=\dfrac{\mu_0\ell_w^2}{4\pi l_0} \).
2. So \( \ell_w=\sqrt{\dfrac{4\pi l_0L}{\mu_0}}=\sqrt{\dfrac{4\pi\times1\times10^{-3}}{4\pi\times10^{-7}}}=100 \) m.

Teacher's Note:
1. The radius cancels out of the final formula.
2. Express N and r through the wire length.

Answer: (d) 85 Wb

Explanation:
1. The bar is a distance \( x=vt=15.6 \) m from the vertex along the bisector. With a 90° angle the half-angle is 45°, so the base is \( 2x \) and the area is \( x^2 \).
2. Flux \( \phi=Bx^2=0.35\times15.6^2\approx85 \) Wb.

Teacher's Note:
1. The area of a right isosceles triangle with height x is x².
2. Use x = vt.

Answer: (d) 57 V

Explanation:
1. \( \phi=Bv^2t^2 \), so \( \varepsilon=\dfrac{d\phi}{dt}=2Bv^2t=2\times0.35\times5.2^2\times3\approx57 \) V.
2. The emf increases linearly with time.

Teacher's Note:
1. Differentiate the flux as a function of time.
2. Compare with \( \varepsilon=B(2x)v \).

Answer: (b) \( \dfrac12B\omega r^2 \)

Explanation:
1. Each element of the rod at distance \( x \) has speed \( \omega x \).
2. Emf \( =\int_0^rB\omega x\,dx=\dfrac12B\omega r^2 \).

Teacher's Note:
1. The rod behaves like a source with emf ½Bωr².
2. Integrate along the rod.