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A closely wound solenoid of 2000 turns and area of cross-section 1.6×10^–4m², carrying a current of 4.0 A, is suspended through its centre allowing it to turn in a horizontal plane.
(a) What is the magnetic moment associated with the solenoid?
(b) What is the torque on the solenoid if a uniform horizontal magnetic field of 7.5 × 10^–2T is set up at an angle of 30⁰ with the axis of the solenoid?

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2.28Am², 0.038 Nm

1.28Am², 0.0038 Nm

1.28Am², 0.048 Nm

1.28Am², 0.042 Nm

answer is B.

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Detailed Solution

M = ni A ; $M = 2000 \times 4 \times 1 .6 \times 10^{- 4} = 1 .28 {Am}^{2}$
$τ = MBsin θ {=1.28\times7.5\times10}^{- 2} \times \frac{1}{2} = 0 .048 Nm$

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