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^-2J. What is the magnitude of the magnetic moment of the magnet?

Answer 1

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