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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:

1. 2√3 J
2. 3 J
3. √3 J
4. 3/2 J

Correct answer: √3 J

Solution

The work done to rotate the needle is given by W = MB(1 - cosθ), where M is the magnetic moment, B is the magnetic field, and θ is the angle. For θ = 60°, cosθ = 1/2, so W = MB(1 - 1/2) = MB/2. Given W = √3 J, we find MB = 2√3. The torque is τ = MBsinθ = (2√3)(√3/2) = 3 J.

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