JEE Physics Electromagnetic Induction MCQs Set 16

Answer: (d) \( -\pi\alpha r^2 \)

Explanation:
1. \( \varepsilon = -\frac{d\phi}{dt} = -\pi r^2\frac{dB}{dt} = -\pi\alpha r^2 \).

Teacher's Note:
1. Flux = \( B \) × area, and the area is \( \pir^{2} \).
2. The constant \( B_{0} \) does not contribute to dB/dt.

Answer: (c) anticlockwise and clockwise

Explanation:
1. On closing \( S \) the flux through \( B \) increases, so \( I_{1} \) opposes it and is opposite to the current in \( A \): anticlockwise.
2. On opening \( S \) the flux decreases, so \( I_{2} \) tries to maintain it and is in the same sense as the original current in \( A \): clockwise.

Teacher's Note:
1. Increasing flux gives an opposing current; decreasing flux gives a supporting current.
2. Steady current produces no induced current.

Answer: (a) n₁ : n₂

Explanation:
1. Peak emf in the coil \( = L\omega I_0 \propto n \) for a given current amplitude, so \( \frac{\varepsilon_1}{\varepsilon_2} = \frac{n_1}{n_2} \).

Teacher's Note:
1. Induced emf = \( L \) dI/dt, and dI/dt ∝ frequency.
2. The current amplitude is the same in both cases.

Answer: (b) 1000 V

Explanation:
1. Energy \( \tfrac12LI^2 = 4 J \). \( \tfrac12CV^2 = 4 \Rightarrow V = \sqrt{\frac{8}{4\times10^{-6}}} \approx 1.4\times10^3 V \).
2. So the rating is of the order of \( 1000 V \).

Teacher's Note:
1. The magnetic energy is transferred to the capacitor.
2. Equate ½LI² with ½CV².

Answer: (a) \( \mu_0ni_0\pi b^2\omega\sin\omega t \)

Explanation:
1. Field inside \( B = \mu_0ni_0\cos\omega t \), flux through the ring \( = B\pi b^2 \).
2. \( \varepsilon = -\frac{d\phi}{dt} = \mu_0ni_0\pi b^2\omega\sin\omega t \).

Teacher's Note:
1. Only the area of the ring itself \( (b \lt a) \) links the field.
2. Differentiate \( \cos \) to get \( \sin \).

Answer: (b) increasing at first and then approaching a constant (terminal) value

Explanation:
1. Eddy currents in the tube oppose the motion, and the opposing force grows with speed.
2. The speed rises until the retarding force equals the weight, after which it is constant.

Teacher's Note:
1. Eddy-current braking is velocity dependent.
2. Terminal speed is reached when magnetic drag equals mg.

Answer: (a) \( \frac{4}{\ln(4/3)} \) s

Explanation:
1. \( I = I_0(1-e^{-t/\tau}) = \frac{I_0}{4} \Rightarrow e^{-t/\tau} = \frac34 \).
2. \( \frac t\tau = \ln\frac43 \Rightarrow \tau = \frac{4}{\ln(4/3)} s \).

Teacher's Note:
1. Growth of current in an LR circuit follows \( 1 - e^{-t/\tau } \).
2. Take the natural \( \log \) of both sides.

Answer: (c) \( \frac\pi2\times10^{-16} \) J

Explanation:
1. \( B = \frac{\mu_0I}{2R} = 2\pi\times10^{-7} T \). Energy density \( u = \frac{B^2}{2\mu_0} = \frac\pi2\times10^{-7} J/m^{3} \).
2. Volume = \( 10^{-9} m^{3} \), so \( U = \frac\pi2\times10^{-16} J \).

Teacher's Note:
1. Energy density of a magnetic field is \( B^{2}/2\mu_{0} \).
2. 1 mm³ = \( 10^{-9} m^{3} \).