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Q.

A magnetic needle suspended parallel to a  magnetic field requires 3J of work to turn it through 60°. The torque needed to maintain the needle in this position will be:

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a

3J

b

3J

c

32J

d

23J

answer is B.

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Detailed Solution

W=−mBcos⁡θ2−cos⁡θ1=−mBcos⁡60∘−cos⁡0∘=−mB12−1=mB2=3Jτ=mBsin⁡60∘=mB×32=mB2×3=3×3=3J.

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