Magnetic Effects & Magnetism

JEE Main level, numerical answer, medium

184 of 276

Question 184 of 276

A circular coil of radius 22 cm and 125125 turns carries a current of 11 A. The coil is placed in a uniform magnetic field of magnitude 0.40.4 T. The axis of the coil makes an angle of 30°30° with the direction of the magnetic field. The torque acting on the coil is α×10−4\alpha \times 10^{-4} N.m. The value of α\alpha is ______.
(π=3.14\pi=3.14)

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Answer: 314

The magnetic moment of the coil is given by M=NIAM = NIA.

The torque acting on the coil is τ=MBsin⁡θ=NIABsin⁡θ\tau = MB \sin\theta = NIAB \sin\theta.

Substituting the given values:

τ=125×1×(π×(0.02)2)×0.4×sin⁡30∘\tau = 125 \times 1 \times (\pi \times (0.02)^2) \times 0.4 \times \sin 30^\circ

τ=125×1×4π×10−4×0.4×0.5\tau = 125 \times 1 \times 4\pi \times 10^{-4} \times 0.4 \times 0.5

τ=100π×10−4\tau = 100\pi \times 10^{-4}

τ=100×3.14×10−4=314×10−4\tau = 100 \times 3.14 \times 10^{-4} = 314 \times 10^{-4} N.m

Comparing with α×10−4\alpha \times 10^{-4} N.m, we get α=314\alpha = 314.

Answer: 314314

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