A magnetic needle suspended parallel to a magnetic field requires √(3)J of work to turn it through 60°. The torque needed to maintain the needle in this position will be (a) 2√(3)J (b) 3 J (c) √(3) J (d) 3/2 J

Answer: (b). W = MB(1-cos 60°) = MB/2 = √(3) J ⇒ MB = 2√(3) J τ = MBsin 60° = 2√(3) × √(3)/2 = 3 So the answer is (b) 3 J (i.e. 3 N m).

A magnetic needle suspended parallel to a magnetic field requires… | NEET 2012 (AIPMT, Mains)

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Field of a Magnet & Magnet in a Uniform Field

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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Answer: (b)

$W = MB(1-\cos 60^\circ) = \frac{MB}{2} = \sqrt{3}\ \text{J} \Rightarrow MB = 2\sqrt{3}\ \text{J}$
$\tau = MB\sin 60^\circ = 2\sqrt{3} \times \frac{\sqrt{3}}{2} = 3$
So the answer is (b) $3\ \text{J}$ (i.e. 3 N m).

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