JEE Main 2026 Apr 2 Shift 1, Physics Q13: Faraday’s Law of Induction

JEE Main2026Apr 2, Shift 1Physics

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 coil number of turns, area of the coil and radius of the coil wire are $2 N$ , $2 A$ and $3 r$ 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
$36$
B
$1282$
C
$16$
D
$64$

Solution

The induced emf has magnitude $ε = N A ∣ d B / d t ∣$ and the coil resistance is $R = \frac{ρ ℓ}{π r ^{2}}$ , where $ℓ = N (2 π R_{coil})$ and $R_{coil} = A / π$ .

So $R \propto \frac{N A}{r ^{2}}$ and the power dissipated $P = \frac{ε ^{2}}{R} \propto \frac{N ^{2} A ^{2}}{N A / r ^{2}} = N A^{3/2} r^{2}$ .

Forming the ratio of powers for the two coils:

$\frac{P ^{'}}{P} = \frac{( 2 N ) ( 2 A ) ^{3/2} ( 3 r ) ^{2}}{N \cdot A ^{3/2} \cdot r ^{2}} = 2 \times 22 \times 9 = 362$

Setting $P^{'} = 2 α P$ gives $α = 36$ .

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JEE Main 2026 Apr 2 Shift 1, Physics Q13: Faraday’s Law of Induction

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