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

Question

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Correct answer: 0

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

A circular coil of radius 2 cm and 125 turns carries a current of 1...

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