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Q. A short bar magnet placed with its axis at $30^{\circ}$ with an external field of $800 \,G$ experiences a torque of $0.016\, Nm$. The magnetic moment of the magnet is, the work done in moving it from its most stable to most unstable position is that

Magnetism and Matter

Solution:

The most stable position is $\theta=0^{\circ}$ and the most unstable position is $\theta=180^{\circ}$, work done is given by
$W =U_{m}\left(\theta=180^{\circ}\right)-U_{m}\left(\theta=0^{\circ}\right)$
$\Rightarrow =2 m B=2 \times 0.40 \times 800 \times 10^{-4}$
$=0.064\, J$