A magnetic needle lying parallel to a magnetic field requires W units of work to turn it through 60^ . The torque needed to maintain the needle in thi - Physics MCQ AIEEE 2003 2026 | ExamTest.live

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A magnetic needle lying parallel to a magnetic field requires W units of work to turn it through 6 0^{\circ} . The torque needed to maintain the needle in this position will be:

Solution & Step-by-step Explanation

Work done in rotating a magnet from θ_{1} to θ_{2} is: W = - MB (cos θ_{2} - cos θ_{1}) Given θ_{1} = 0^{\circ} and θ_{2} = 6 0^{\circ} : W = - MB (cos 6 0^{\circ} - cos 0^{\circ}) = - MB (1/2 - 1) = \frac{MB}{2} ⟹ MB = 2 W The torque required at θ = 6 0^{\circ} is: τ = MB sin θ = (2 W) sin 6 0^{\circ} = 2 W \times \frac{3}{2} = 3 W

Try it yourself before checking the explanation above.

A magnetic needle lying parallel to a magnetic field requires W units of work to turn it through 6 0^{\circ} . The torque needed to maintain the needle in this position will be:

3 W

W

(3 /2) W

2 W

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