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 10-2J. What is the magnitude of the magnetic moment of the magnet?
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θ = 30°, B = 0.25 T, τ = 4.5 x 10-2J, m = ?, τ = mBsinθ ∴ m = \(\frac { τ }{ mBsinθ }\) = \(\frac{4.5 \times 10^{-2}}{0.25 \times \sin 30^{\circ}}\) = \(\frac{4.5 \times 10^{-2} \times 2}{0.25}\) = 36.0 x 10-2 = 0.36 Am2 (or) JT-1
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