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

A conducting square loop of side

L

, mass

M

and resistance

R

is moving in the

X Y

plane with its edges parallel to the

X

and

Y

axes. The region

y \geq 0

has a uniform magnetic field,

B = B_{0} k

. The magnetic field is zero everywhere else. At time

t = 0

, the loop starts to enter the magnetic field with an initial velocity

v_{0} \hat{J} m / s

, as shown in the figure. Considering the quantity

K = \frac{B _{0}^{2} L ^{2}}{R M}

in appropriate units, ignoring self-inductance of the loop and gravity, which of the following statements is/are correct:

(A)If $v_{0} = 1.5 K L$ , the loop will stop before it enters completely inside the region of magnetic field.
(B)When the complete loop is inside the region of magnetic field, the net force acting on the loop is zero.✓ Correct
(C)If $v_{0} = \frac{K L}{10}$ , the loop comes to rest at $t = (\frac{1}{K}) ln (\frac{5}{2})$ .
(D)If $v_{0} = 3 K L$ , the complete loop enters inside the region of magnetic field at time $t = (\frac{1}{K}) ln (\frac{3}{2})$ .✓ Correct

Answer: (B), (D)

Solution

The induced emf is

ε = B_{0} v L

, giving current

i = \frac{B _{0} v L}{R}

. The magnetic force is

F = B_{0} i L = \frac{B _{0}^{2} L ^{2} v}{R}

, opposing motion.

This gives acceleration

a = - K v

, where

K = \frac{B _{0}^{2} L ^{2}}{R M}

.

Solving

\frac{d v}{d t} = - K v

gives

v = v_{0} e^{- K t}

.

Integrating:

x = \frac{v _{0}}{K} (1 - e^{- K t})

.

(B) When the loop is completely inside,

\frac{d ϕ}{d t} = 0

, so no induced emf, no current, and no force. ✓

(D) For

v_{0} = 3 K L

, when

x = L

L = \frac{3 K L}{K} (1 - e^{- K t}) \Rightarrow \frac{1}{3} = 1 - e^{- K t} \Rightarrow e^{- K t} = \frac{2}{3}

t = \frac{1}{K} ln (\frac{3}{2})

✓

JEE Advanced · 2025 · 18 May 2025 · Paper 1 · One or more correct options