JEE Main 2026 Apr 5 Shift 2, Physics Q25: Faraday’s Law of Induction

JEE Main2026Apr 5, Shift 2Physics

A circular loop of radius 20 cm and resistance 2 $Ω$ is placed in a time varying magnetic field $B = (2 t^{2} + 2 t + 3)$ T. At $t = 0$ , for the plane of the loop being perpendicular to the magnetic field, the induced current in the loop at $t = 3$ s is $\frac{α}{50}$ A. The value of $α$ is _____. (Take $π = 22/7$ )

Your answer

Solution

Flux: $ϕ = B \cdot π r^{2}$ . EMF: $ε = π r^{2} \frac{d B}{d t}$ .

$\frac{d B}{d t} = 4 t + 2$ . At $t = 3$ : $\frac{d B}{d t} = 14$ T/s.

$ε = π (0.2)^{2} \times 14 = \frac{22}{7} \times 0.04 \times 14 = \frac{22 \times 0.56}{7} = \frac{12.32}{7} = 1.76$ V.

$i = \frac{ε}{R} = \frac{1.76}{2} = 0.88 = \frac{44}{50}$ A.

$∴ α = 44$ .

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JEE Main 2026 Apr 5 Shift 2, Physics Q25: Faraday’s Law of Induction

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