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 to
4.5 x 10-2 J. What is the magnitude of magnetic moment of the magnet?​

Answer 1

Answer:

Given data

angle between magnetic moment and magnetic field θ=30

0

External magetic field B=0.25T

Torque τ=4.5×10

−2

J

The torque experienced by a bar magnet placed at an angle θ with the external magnetic field B is given by

τ=MBsinθ

⟹4.5×10

−2

=M×0.25sin30

∘

⟹M=0.36JT

−1

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