JEE Physics Electromagnetic Induction MCQs Set 22

Answer: (c) \( 3\times10^{-5}\ \text s,\ 10\ \text A \)

Explanation:
1. Each half has \( L'=0.9\times10^{-4} \) H and \( R'=3\ \Omega \). In parallel: \( L_{eq}=0.45\times10^{-4} \) H, \( R_{eq}=1.5\ \Omega \).
2. \( \tau=\dfrac{L_{eq}}{R_{eq}}=3\times10^{-5} \) s and \( I=\dfrac{15}{1.5}=10 \) A.

Teacher's Note:
1. Write Faraday's law (or the circuit equation) first and identify what is changing, flux, area, field or current.
2. Use Lenz's law only to fix the direction, then check the limiting case.

Answer: (a) 3.465 s

Explanation:
1. Energy \( \propto i^2 \): one quarter of the maximum means \( i=\dfrac{I_0}2 \). \( \tau=\dfrac LR=5 \) s.
2. \( 1-e^{-t/\tau}=\dfrac12\Rightarrow t=\tau\ln2=5\times0.693=3.465 \) s.

Teacher's Note:
1. Write Faraday's law (or the circuit equation) first and identify what is changing, flux, area, field or current.
2. Use Lenz's law only to fix the direction, then check the limiting case.

Answer: (b) 12 mV

Explanation:
1. \( \vec v\times\vec B=(6\hat j)\times(0.1\hat k)=0.6\hat i \): the induced field points along x.
2. \( V=vBl=6\times0.1\times0.02=12\times10^{-3} \) V.

Teacher's Note:
1. Write Faraday's law (or the circuit equation) first and identify what is changing, flux, area, field or current.
2. Use Lenz's law only to fix the direction, then check the limiting case.

Answer: (d) zero

Explanation:
1. The flux through the cylinder's cross-section is constant (B is uniform and steady).
2. No emf is induced, so there is no induced current.

Teacher's Note:
1. Write Faraday's law (or the circuit equation) first and identify what is changing, flux, area, field or current.
2. Use Lenz's law only to fix the direction, then check the limiting case.

Answer: (a) the same

Explanation:
1. EMF \( \propto N \): it becomes 4 times. Resistance \( \propto\dfrac{N}{r_w^2} \) (length \( \propto N \), area \( \propto r_w^2 \)): it becomes \( 4\times4=16 \) times.
2. Power \( =\dfrac{\varepsilon^2}{R}\propto\dfrac{16}{16}=1 \): unchanged.

Teacher's Note:
1. Write Faraday's law (or the circuit equation) first and identify what is changing, flux, area, field or current.
2. Use Lenz's law only to fix the direction, then check the limiting case.

Answer: (c) passes through a maximum

Explanation:
1. \( B(t)\propto I_1(t) \) rises from zero to a constant, while \( I_2\propto-\dfrac{dI_1}{dt} \) starts at a maximum and decays to zero.
2. The product is zero at \( t=0 \) (B = 0) and zero as \( t\to\infty \) (\( I_2=0 \)), so it rises and then falls: it passes through a maximum.

Teacher's Note:
1. Write Faraday's law (or the circuit equation) first and identify what is changing, flux, area, field or current.
2. Use Lenz's law only to fix the direction, then check the limiting case.

Answer: (d) is repelled by the loop A

Explanation:
1. The increasing current of A increases the flux through B.
2. By Lenz's law B's induced current is opposite to A's, and opposite currents in neighbouring coplanar loops repel.

Teacher's Note:
1. Write Faraday's law (or the circuit equation) first and identify what is changing, flux, area, field or current.
2. Use Lenz's law only to fix the direction, then check the limiting case.

Answer: (b) \( \dfrac{BR}{2} \)

Explanation:
1. \( \oint\vec E\cdot d\vec l=\left|\dfrac{d\Phi}{dt}\right| \): \( E\cdot2\pi R=\pi R^2\dfrac{dB}{dt}=\pi R^2B \) (since B rises by B in 1 s).
2. \( E=\dfrac{BR}{2} \).

Teacher's Note:
1. Write Faraday's law (or the circuit equation) first and identify what is changing, flux, area, field or current.
2. Use Lenz's law only to fix the direction, then check the limiting case.

Answer: (c) a decrease of magnitude \( \dfrac{\gamma BQR^2}{2} \)

Explanation:
1. Torque on the charge: \( \tau=QER=\dfrac{BQR^2}{2} \). The induced field opposes the increasing flux, i.e. acts against the circulation (anticlockwise motion), so the angular momentum decreases.
2. \( \Delta L=\tau\,\Delta t=\dfrac{BQR^2}{2}\times1 \), so \( |\Delta\mu|=\gamma\dfrac{BQR^2}2 \), a decrease.

Teacher's Note:
1. Write Faraday's law (or the circuit equation) first and identify what is changing, flux, area, field or current.
2. Use Lenz's law only to fix the direction, then check the limiting case.

Answer: (a) The magnetic force

Explanation:
1. Induced currents in the track coils interact with the magnetic field of the train's magnets.
2. The resulting force on current-carrying conductors in a magnetic field is the magnetic force.

Teacher's Note:
1. Write Faraday's law (or the circuit equation) first and identify what is changing, flux, area, field or current.
2. Use Lenz's law only to fix the direction, then check the limiting case.

Answer: (b) \( \rho_A\lt \rho_B \) and \( m_A=m_B \)

Explanation:
1. The induced emf is the same in both rings; the lower resistivity ring carries more current and feels a larger repulsive force.
2. For the same mass, \( h_A\gt h_B \) needs a larger force on A, i.e. \( \rho_A\lt \rho_B \). A larger mass of A with equal or higher resistivity would make it jump lower.

Teacher's Note:
1. Write Faraday's law (or the circuit equation) first and identify what is changing, flux, area, field or current.
2. Use Lenz's law only to fix the direction, then check the limiting case.

Answer: (d) 6

Explanation:
1. Field inside the tube (like a solenoid with one turn per 10 m): \( B=\dfrac{\mu_0I}{10} \). Flux through the loop: \( \Phi=B\pi r^2=\dfrac{\mu_0I_0\cos300t}{10}\pi(0.1)^2 \).
2. \( \varepsilon=300\times\dfrac{\mu_0I_0\pi\times10^{-2}}{10}\sin300t \), \( i=\dfrac\varepsilon{0.005}=60\pi\mu_0I_0\sin300t \). Moment \( =i\pi r^2=60\pi\times0.01\pi\,\mu_0I_0\sin300t=0.6\pi^2\mu_0I_0\sin300t\approx5.9\,\mu_0I_0\sin300t \), so N = 6.

Teacher's Note:
1. Write Faraday's law (or the circuit equation) first and identify what is changing, flux, area, field or current.
2. Use Lenz's law only to fix the direction, then check the limiting case.

Answer: (a) \( \dfrac{\mu_0Ldna^2i_0\omega\cos\omega t}{2\rho R} \)

Explanation:
1. Flux through the shell \( =\mu_0ni\,\pi a^2 \), so \( \varepsilon=\mu_0n\pi a^2i_0\omega\cos\omega t \).
2. Resistance of the shell round its circumference: \( R_s=\rho\dfrac{2\pi R}{Ld} \). Current \( =\dfrac{\varepsilon}{R_s}=\dfrac{\mu_0Ldna^2i_0\omega\cos\omega t}{2\rho R} \).

Teacher's Note:
1. Write Faraday's law (or the circuit equation) first and identify what is changing, flux, area, field or current.
2. Use Lenz's law only to fix the direction, then check the limiting case.

Answer: (b) \( \dfrac94\dfrac{\mu_0^2M^2a^4v}{Rx^8} \)

Explanation:
1. Axial field of the dipole: \( B=\dfrac{\mu_0}{4\pi}\dfrac{2M}{x^3} \). Flux through the small loop: \( \Phi=B\pi a^2=\dfrac{\mu_0Ma^2}{2x^3} \).
2. \( \varepsilon=\left|\dfrac{d\Phi}{dt}\right|=\dfrac{3\mu_0Ma^2v}{2x^4} \). Power \( =\dfrac{\varepsilon^2}{R}=\dfrac{9\mu_0^2M^2a^4v^2}{4Rx^8} \), and the force \( F=\dfrac{P}{v}=\dfrac94\dfrac{\mu_0^2M^2a^4v}{Rx^8} \).

Teacher's Note:
1. Write Faraday's law (or the circuit equation) first and identify what is changing, flux, area, field or current.
2. Use Lenz's law only to fix the direction, then check the limiting case.