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:

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

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:

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

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$\sqrt{3} J$

$\frac{3}{2} J$

$2 \sqrt{3} J$

answer is B.

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Detailed Solution

$\begin{array}{l} W = - mB (\cos θ_{2} - \cos θ_{1}) = - mB (\cos 60^{\circ} - \cos 0^{\circ}) = - mB (\frac{1}{2} - 1) = \frac{mB}{2} = \sqrt{3} J \\ τ = mBsin 60^{\circ} = mB \times \frac{\sqrt{3}}{2} = \frac{mB}{2} \times \sqrt{3} = \sqrt{3} \times \sqrt{3} = 3 J . \end{array}$