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=0.25 T , 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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