A short bar magnet placed with its axis at 30^ with an external field of 800 Gauss, experiences a torque of 0.016 ~N. m . The work done in moving it from most stable to most unstable position is α × 10^ -3 ~J . The

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For a magnetic dipole in an external field, torque is

\tau = mB\sin\theta

At 30°:

0.016 = m \times 0.08 \times \sin 30°

0.016 = m \times 0.08 \times 0.5

m = 0.4

A·m

^2

Work done from most stable (aligned, 0°) to most unstable (anti-aligned, 180°) position equals the change in potential energy:

W = U_0° - U_{180°} = -mB\cos 0° - (-mB\cos 180°) = -mB - mB = 2mB

W = 2 \times 0.4 \times 0.08 = 0.064

J

= 64 \times 10^{-3}

J
Therefore,

\alpha = 64

A short bar magnet placed with its axis at 30^ with an external...

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