JEE Main · 2021 · 16 Mar 2021 · Shift 2 · Electromagnetic Induction

The magnetic field in a region is given by

B = B_{0} (\frac{x}{a}) \hat{k}

. A square loop of side d is placed with its edges along the x and y axes. The loop is moved with a constant velocity

v = v_{0} \hat{i}

. The emf induced in the loop is:

Answer: (A)

Solution

Let the near edge of the loop be at

x

. Since the field varies across the loop, the flux is found by integrating over strips of width

d x^{'}

and length

d

Φ = \int_{x}^{x + d} B_{0} \frac{x ^{'}}{a} d d x^{'} = \frac{B _{0} d}{2 a} [(x + d)^{2} - x^{2}] = \frac{B _{0} d}{2 a} (2 x d + d^{2})

The loop moves along

x

with

\frac{d x}{d t} = v_{0}

, and

d

and

a

are constants, so

\frac{d Φ}{d t} = \frac{B _{0} d}{2 a} \cdot 2 d v_{0} = \frac{B _{0} v _{0} d ^{2}}{a}

The emf induced in the loop is

\frac{B _{0} v _{0} d ^{2}}{a}

JEE Main · 2021 · 16 Mar 2021 · Shift 2 · Single correct option