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JEE Mains · Physics · STD 12 - Moving charges and magnetism
  • JEE Mains 2026
  • Medium
  • Numerical

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 \(\alpha \times 10^{-4}\) N.m. The value of \(\alpha\) is ______.
(\(\pi=3.14\))

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Verified Solution

Answer & Solution

Correct Answer

(B) 314

Step-by-step Solution

Detailed explanation

The magnetic moment of the coil is given by \(M = NIA\).

The torque acting on the coil is \(\tau = MB \sin\theta = NIAB \sin\theta\).

Substituting the given values:

\(\tau = 125 \times 1 \times (\pi \times (0.02)^2) \times 0.4 \times \sin 30^\circ\)

\(\tau = 125 \times 1 \times 4\pi \times 10^{-4} \times 0.4 \times 0.5\)

\(\tau = 100\pi \times 10^{-4}\)

\(\tau = 100 \times 3.14 \times 10^{-4} = 314 \times 10^{-4}\) N.m

Comparing with \(\alpha \times 10^{-4}\) N.m, we get \(\alpha = 314\).

Answer: \(314\)
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