A short bar magnet placed with its axis at 30° with a uniform external magnetic field of 0.5 T experiences a torque of magnitude equal to 4.5 × 10⁻² J. Then the magnitude of magnetic moment of the magnet will be .

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(4) 18 × 10⁻² JT⁻¹. Given: θ=30°, B=0.5 T, τ=4.5×10⁻² J (a torque in J per radian is the same as one in N m). Idea: a magnetic dipole in a uniform field feels a torque τ=mB sin θ, where θ is the angle between the magnet's axis and the field. Rearranged: m=τ/(B sin θ). sin 30°=0.5, so B sin θ=0.5×0.5=0.25 T. m=(4.5×10⁻²)/(0.25)=0.18 J T⁻¹, that is 18×10⁻² J T⁻¹.

A short bar magnet placed with its axis at 30° with a uniform external magnetic field of 0.5 T experiences a torque of magnitude equal to 4.5 × 10⁻² J. Then the magnitude of magnetic moment of the magnet will be ________.

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