JEE Physics Electromagnetic Induction MCQs Set 20

Answer: (b) 437.5 J

Explanation:
1. \( \varepsilon=L\dfrac{dI}{dt}\Rightarrow L=\dfrac{25}{15}=\dfrac53 \) H.
2. \( \Delta U=\tfrac12L(I_2^2-I_1^2)=\tfrac12\times\dfrac53\times(625-100)=437.5 \) J.

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) 250 Wb

Explanation:
1. Charge flowed \( q=\int I\,dt=\) area of the I-t graph \( =\tfrac12\times10\times0.5=2.5 \) C.
2. \( q=\dfrac{\Delta\Phi}{R}\Rightarrow\Delta\Phi=qR=2.5\times100=250 \) Wb.

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) zero

Explanation:
1. Magnetic field lines are closed curves: the lines passing up through the inside of the loop return downwards through the outside part of the same plane.
2. The net flux through the whole infinite plane is therefore zero (Gauss's law for magnetism).

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) left to right, as the induced current opposes the decrease of current

Explanation:
1. The current is decaying, so the flux linked with the wire's own inductance decreases.
2. By Lenz's law the induced emf opposes the decrease, so the induced current is in the same direction as the original current: left to right.

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) in the direction opposite to the current in A

Explanation:
1. As A approaches B, the flux of A through B increases.
2. By Lenz's law B's induced current produces a field opposing this increase, so it circulates opposite to the current in A (coil C is fixed, so its contribution is constant).

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) clockwise

Explanation:
1. Above a wire carrying current to the right, the field points out of the paper, and it increases with the current.
2. The induced current must produce a field into the paper, which needs a clockwise 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: (c) \( \dfrac{5B\omega l^2}{2} \)

Explanation:
1. The rod occupies radii from 2l to 3l about the fixed point.
2. \( \varepsilon=\int_{2l}^{3l}B\omega r\,dr=\dfrac{B\omega}{2}\left(9l^2-4l^2\right)=\dfrac{5B\omega l^2}{2} \).

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: (d) in both AD and BC

Explanation:
1. Each side perpendicular to v is a moving conductor in which charges experience \( q\vec v\times\vec B \).
2. So an electric field \( E=vB \) exists along both AD and BC. The emfs cancel round the loop, so no current flows, but the fields are there in both sides.

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) \( 2RBv \), with Q at higher potential

Explanation:
1. The effective length of a curved conductor moving in a uniform field is the chord joining its ends: \( l=2R \) (perpendicular to v and B).
2. \( \varepsilon=B(2R)v \). With \( \vec v \) downwards and \( \vec B \) into the page, \( \vec v\times\vec B \) points to the right, pushing positive charge towards Q.

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) zero

Explanation:
1. The flux \( BL^2 \) through the loop does not change as it translates in a uniform field.
2. The emfs in the two opposite sides cancel, so the net emf and the current are zero.

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) \( \dfrac{mgR}{B^2L^2} \)

Explanation:
1. Terminal: the magnetic force on the rod equals the weight of the hanging mass (the rod is massless): \( BIL=mg \).
2. \( I=\dfrac{BLv}{R}\Rightarrow\dfrac{B^2L^2v_T}{R}=mg\Rightarrow v_T=\dfrac{mgR}{B^2L^2} \).

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: (b) \( \dfrac g2 \)

Explanation:
1. The magnetic drag is proportional to v. At \( v_T/2 \) it equals \( mg/2 \).
2. \( a=\dfrac{mg-mg/2}{m}=\dfrac g2 \).

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) \( \dfrac{B\omega r^2}{2R}\left[1-e^{-(R/L)t}\right] \)

Explanation:
1. A constant emf \( \varepsilon \) drives an L-R circuit: \( i=\dfrac{\varepsilon}{R}\left(1-e^{-Rt/L}\right) \).
2. With \( \varepsilon=\dfrac{B\omega r^2}{2} \): \( i=\dfrac{B\omega r^2}{2R}\left[1-e^{-(R/L)t}\right] \).

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) 1 m/s

Explanation:
1. At terminal speed the power supplied by gravity equals the total electrical power: \( mgv=0.76+1.2=1.96 \) W.
2. \( v=\dfrac{1.96}{0.2\times9.8}=1 \) m/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.