JEE Main 2026 Apr 4 Shift 2, Physics Q24: Torque and Work Done in a Magnetic Field

JEE Main2026Apr 4, Shift 2Physics

A circular coil of radius 2 cm and 125 turns carries a current of 1 A. The coil is placed in a uniform magnetic field of magnitude 0.4 T. The axis of the coil makes an angle of 30° with the direction of the magnetic field. The torque acting on the coil is $α \times 1 0^{- 4} N.m$ . The value of $α$ is ________. ( $π = 3.14$ )

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Solution

Torque on a current loop: $τ = ∣ M \times B ∣ = N I A B sin θ$ , where $θ$ is the angle between the coil's magnetic moment (along the axis) and $B$ .

$A = π r^{2} = π (0.02)^{2} m^{2}$ .

$τ = 125 \times 1 \times π (0.02)^{2} \times 0.4 \times sin 30°$

$= 125 \times π \times 4 \times 1 0^{- 4} \times 0.4 \times 0.5 = 100 π \times 1 0^{- 4} N.m$ .

$α = 100 π = 100 \times 3.14 = 314$ .

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JEE Main 2026 Apr 4 Shift 2, Physics Q24: Torque and Work Done in a Magnetic Field

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