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Question

A short bar magnet placed with its axis at 30° with a uniform external magnetic field of 0.25 T experiences a torque of magnitude equal to4.5 × 10–2 J. What is the magnitude of magnetic moment of the magnet?

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Solution

Given: The angle between magnetic field and magnetic moment is 30°, magnitude of the external magnetic field B=0.25T , and the torque experienced by the magnet is 4.5× 10 2 J.

The torque is given as,

T=MBsinθ

Where, M is the magnitude of the magnetic moment.

By substituting the given values in the above equation, we get.

4.5× 10 2 =M( 0.25sin30° ) M=0.36 JT 1

Thus, magnitude of the magnetic moment is 0.36 JT 1 .


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