JEE Physics Electromagnetic Induction MCQs Set 08

Answer: (c) 4B0L² Wb

Explanation:
1. The area vector is along k̂, so only the z-component of B contributes.
2. \( \Phi=\vec B\cdot\vec A=4B_0L^2\ \text{Wb} \)

Teacher's Note:
1. Flux is B.A, so only the component of B along the area vector counts.

Answer: (d) Zero

Explanation:
1. Only the z-component of B passes through the loop (area vector along z).
2. B_z = 50 T is constant, so the flux does not change and the emf is zero.

Teacher's Note:
1. Induced emf needs a changing flux through the loop; the changing components lie in the plane of the loop.

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

Explanation:
1. \( I=\frac{E}{R}\left(1-e^{-Rt/L}\right)=1\cdot\left(1-e^{-5\times2/10}\right)=1-e^{-1} \)

Teacher's Note:
1. Growth of current in an LR circuit has time constant L/R = 2 s.

Answer: (c) High inductance and low resistance

Explanation:
1. A choke reduces AC current through its large inductive reactance XL = ωL.
2. Low resistance keeps the power loss small.

Teacher's Note:
1. A choke controls AC without wasting power, so it needs high L and low R.

Answer: (c) \( \frac{2}{\ln(10/9)} \)

Explanation:
1. \( 18=20e^{-2/\tau}\Rightarrow\frac{2}{\tau}=\ln\frac{10}{9} \)
2. \( \tau=\frac{2}{\ln(10/9)} \)

Teacher's Note:
1. Decay of current is exponential: I = I0 e^(−t/τ).

Answer: (b) l decreases and A increases

Explanation:
1. \( L=\frac{\mu_0N^2A}{l} \) with N fixed.
2. L grows when A increases and when l decreases.

Teacher's Note:
1. Self inductance is proportional to the area and inversely proportional to the length.

Answer: (d) A is false, but R is true.

Explanation:
1. The iron core raises the inductance, so the reactance XL = ωL rises (reason true).
2. A larger XL reduces the current, so the bulb glows dimmer and the assertion is false.

Teacher's Note:
1. More inductance means more opposition to AC and less bulb current.

Answer: (b) \( \frac{(B\pi r^2\omega)^2}{8R} \)

Explanation:
1. Flux \( \Phi=\frac{B\pi r^2}{2}\cos\omega t \), so peak emf \( \varepsilon_0=\frac{B\pi r^2\omega}{2} \).
2. \( \bar P=\frac{\varepsilon_0^2}{2R}=\frac{(B\pi r^2\omega)^2}{8R} \)

Teacher's Note:
1. Mean power of a sinusoidal emf is ε0²/2R.

Answer: (d) \( 2.4\pi\times10^{-4}\ \text{H} \)

Explanation:
1. \( M=\frac{\mu_0N_1N_2A}{l}=\frac{4\pi\times10^{-7}\times300\times400\times10^{-3}}{0.2} \)
2. \( =2.4\pi\times10^{-4}\ \text{H} \)

Teacher's Note:
1. Mutual inductance of coaxial solenoids is μ0N1N2A/l.

Answer: (d) VB = 45 mV; VW = 120 mV with left side of pilot at higher voltage

Explanation:
1. Vertical component \( B_V=B\sin\theta=3.33\times10^{-5}\ \text{T} \); horizontal component \( B_H=B\cos\theta=3.73\times10^{-5}\ \text{T} \).
2. Wing tips: \( V_W=B_Vv\ell=3.33\times10^{-5}\times240\times15=120\ \text{mV} \). Top-bottom: \( V_B=B_Hv h=3.73\times10^{-5}\times240\times5\approx45\ \text{mV} \).
3. In the northern hemisphere, flying east, the force on positive charges is towards the north, which is the pilot's left, so the left side is at higher potential.

Teacher's Note:
1. Wing span cuts the vertical component of the field; height cuts the horizontal component.

Answer: (b) 16

Explanation:
1. \( \frac{dU}{dt}=Li\frac{di}{dt}=2\times2\times4=16\ \text{J/s} \)

Teacher's Note:
1. Differentiate U = ½Li² with respect to time.

Answer: (c) 512

Explanation:
1. \( L\frac{di}{dt}=4t\Rightarrow i=2t^2=32\ \text{A at }t=4\ \text{s} \)
2. \( U=\frac12Li^2=\frac12(1)(32)^2=512\ \text{J} \)

Teacher's Note:
1. Integrate V/L to get the current, then use U = ½Li².

Answer: (a) 10

Explanation:
1. \( \varepsilon=\left|\frac{d\phi}{dt}\right|=|20t-50|=|60-50|=10\ \text{V} \)

Teacher's Note:
1. Faraday's law: emf is the rate of change of flux.

Answer: (b) 1 ms

Explanation:
1. Final current \( I_0=\frac{12}{6}=2\ \text{A} \), time constant \( \tau=\frac LR=1.4\ \text{ms} \).
2. \( 1=2(1-e^{-t/\tau})\Rightarrow t=\tau\ln2\approx1\ \text{ms} \)

Teacher's Note:
1. Current is half of its final value after τ ln 2.

Answer: (a) 1.67 H

Explanation:
1. \( L=\frac{\varepsilon\,\Delta t}{\Delta i}=\frac{50\times0.1}{3}=1.67\ \text{H} \)

Teacher's Note:
1. Average emf = L (Δi/Δt).

Answer: (b) 6

Explanation:
1. Energy conservation: \( \frac12CV_0^2=\frac12CV^2+\frac12Li^2 \).
2. \( i^2=\frac{C(V_0^2-V^2)}{L}=\frac{2\times10^{-6}(144-36)}{6\times10^{-6}}=36\Rightarrow i=6\ \text{A} \)

Teacher's Note:
1. In an LC circuit the total energy is shared between the capacitor and the inductor.

Answer: (b) 4

Explanation:
1. Induced emf in coil 2: \( \varepsilon=M\frac{dI_1}{dt} \); charge \( q=\frac{\Delta\phi}{R}=\frac{M\Delta I_1}{R} \).
2. \( q=\frac{5\times(8-0)}{10}=4\ \text{C} \)

Teacher's Note:
1. Charge through a closed loop is the change in flux linkage divided by resistance.

Answer: (b) 4

Explanation:
1. Energy stored when the current reaches 2 A: \( U=\frac12Li^2=\frac12(2)(2)^2=4\ \text{J} \)

Teacher's Note:
1. Energy spent equals the energy stored in the magnetic field, which depends only on the final current.

Answer: (b) 104

Explanation:
1. Secondary voltage \( =8\times120=960\ \text{V} \).
2. \( I=\frac{P}{V}=\frac{10^4}{960}\approx10.4\ \text{A}\Rightarrow10I\approx104 \)

Teacher's Note:
1. Power in the load equals secondary voltage times secondary current for an ideal transformer.