JEE Advanced · 2025 · 18 May 2025 · Paper 1 · Electromagnetic Induction

A conducting square loop initially lies in the

X Z

plane with its lower edge hinged along the

X

-axis. Only in the region

y \geq 0

, there is a time dependent magnetic field pointing along the

Z

-direction,

B (t) = B_{0} (cos ω t) \hat{K}

, where

B_{0}

is a constant. The magnetic field is zero everywhere else. At time

t = 0

, the loop starts rotating with constant angular speed

ω

about the

X

axis in the clockwise direction as viewed from the

+ X

axis (as shown in the figure). Ignoring self-inductance of the loop and gravity, which of the following plots correctly represents the induced e.m.f.

(V)

in the loop as a function of time:

Answer: (A)

Solution

The magnetic flux through the loop is:

ϕ = B_{0} cos ω t \cdot A sin ω t = \frac{B _{0} A sin 2 ω t}{2}

This is valid for

0 \leq t \leq \frac{π}{ω}

when the loop is in the region

y \geq 0

.

The induced emf is:

ε = - \frac{d ϕ}{d t} = - B_{0} A ω cos 2 ω t

For

\frac{π}{ω} \leq t \leq \frac{2 π}{ω}

, the loop is in the region

y < 0

where

B = 0

, so

ε = 0

.

This pattern repeats with period

\frac{2 π}{ω}

. Option (A) correctly shows this behavior.

JEE Advanced · 2025 · 18 May 2025 · Paper 1 · Single correct option