JEE Main · 2025 · 04 Apr 2025 · Shift 1 · Electromagnetic Induction

Conductor wire ABCDE with each arm 10 cm in length is placed in magnetic field of

\frac{1}{2}

Tesla, perpendicular to its plane. When conductor is pulled towards right with constant velocity of

10 cm / s

, induced emf between points A and E is
______ mV.

B = \frac{1}{2}

Tesla

Answer: 10

Solution

In a uniform field the bent conductor acts like a straight wire joining its ends, and the induced emf is

ε = B v ℓ_{⊥}

, where

ℓ_{⊥}

is the separation of A and E measured perpendicular to the velocity. Each

10

cm arm meeting at the right angle at C is inclined at

4 5^{\circ}

to the horizontal, so A and E are separated vertically by

2 \times 10 sin 4 5^{\circ} = 102

cm. Hence

ε = \frac{1}{2} \times 0.1 \times 0.1 2 = 0.01 V = 10 mV .

The induced emf between A and E is

10

mV.

JEE Main · 2025 · 04 Apr 2025 · Shift 1 · Integer answer