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 10⁻³ ~J . The value of is _ _ _ _ .

Step-by-step solution

For a magnetic dipole in an external field, torque is = mB At 30°: 0.016 = m 0.08 30° 0.016 = m 0.08 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 = U₀° - U_ 180° = -mB 0° - (-mB 180°) = -mB - mB = 2mB W = 2 0.4 0.08 = 0.064 J = 64 10⁻³ J Therefore, = 64