A short bar magnet placed with its axis at 30° with a uniform external magnetic field of 0.15 T experiences a torque of magnitude equal to 4.5×10⁻² N m. The magnitude of magnetic moment of the magnet is N m T⁻¹.

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(1) 0.60. Given: θ=30°, B=0.15 T, τ=4.5×10⁻² N m. Idea: the torque on a magnetic dipole in a uniform field is τ=mB sin θ, so m=τ/(B sin θ). sin 30°=0.5, so B sin θ=0.15×0.5=0.075 T. m=(4.5×10⁻²)/(0.075)=0.60 N m T⁻¹. The unit N m T⁻¹ is the same as J T⁻¹ and as A m², the three ways of writing a magnetic moment.

A short bar magnet placed with its axis at 30° with a uniform external magnetic field of 0.15 T experiences a torque of magnitude equal to 4.5×10⁻² N m. The magnitude of magnetic moment of the magnet is ______ N m T⁻¹.

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