JEE Main · 2026 · 04 Apr 2026 · Shift 2 · Magnetic Effects of Current

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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

3 0^{\circ}

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

)

Answer: 314

Solution

The torque on the coil is

τ = N I A B sin θ

, where

θ

is the angle between the coil's axis (its magnetic moment) and the field. With

A = π r^{2} = π (0.02)^{2} = 4 π \times 1 0^{- 4} m^{2}

and

θ = 3 0^{\circ}

τ = 125 \times 1 \times (4 π \times 1 0^{- 4}) \times 0.4 \times \frac{1}{2} = 100 π \times 1 0^{- 4} = 314 \times 1 0^{- 4} N m

(taking

π = 3.14

). The value of

α

314

JEE Main · 2026 · 04 Apr 2026 · Shift 2 · Integer answer

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