A short bar magnet with its axis at 30 degree to an external magnetic field of 800G experience a torque of 0.016Nm. Find the (i) magnetic moment of magnet (ii) Work done in turning the dipole from its stable to unstable position.

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Solution

T = m B sin30

0.016 N = m 800 10^-4(1/2)
m= 0.0162/810^-2.
m= 1.6*2/8= 0.4 Am².

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