JEE Main · 2026 · 02 Apr 2026 · Shift 1 · Electromagnetic Induction

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When a coil is placed in a time dependent magnetic field the power dissipated in it is P. The number of turns, area of the coil and radius of the coil wire are N, A and r respectively. For a second coils number of turns, area of the coil and radius of the coil wire are 2N, 2A and 3r respectively. When the first coil is replaced with second coil the power dissipated in it is

2 α P

. The value of

α

is ____.

(A)36✓ Correct
(B) $1282$
(C)16
(D)64

Answer: (A)

Solution

The power dissipated is

P = \frac{ε ^{2}}{R}

, where the induced emf

ε = N A \frac{d B}{d t}

and the coil resistance

R = \frac{ρ ℓ}{π r ^{2}}

. The wire length is

ℓ = N (2 π a)

with the loop radius

a = A / π

from

A = π a^{2}

, so

R \propto \frac{N A}{r ^{2}}

. Therefore

P \propto \frac{( N A ) ^{2}}{N A / r ^{2}} = N A^{3/2} r^{2}

. Replacing

N, A, r

2 N, 2 A, 3 r

multiplies this by

2 \cdot 2^{3/2} \cdot 9 = 362

, so

P^{'} = 362 P = 2 (36) P

. The value of

α

36

JEE Main · 2026 · 02 Apr 2026 · Shift 1 · Single correct option

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JEE Main · 2026 · 02 Apr 2026 · Shift 1 · Electromagnetic Induction

Solution

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