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A short bar magnet is placed in an external magnetic field of 600 G. When its axis makes an angle of 30° with the external field, it experiences a torque of 0.012 Nm. What is the magnetic moment of the magnet ? |
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Answer» `0.2 "Am"^(2)` `THEREFORE m- (tau)/( B sin theta) = (0.012)/(600 XX 10^(-4)xx sin30^(@) )` `therefore m= (0.012)/(6 xx 10^(-2)xx (1)/(2) ) therefore m= 0.4 "Am"^(2)` |
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