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A magnet is suspendedat an angle60^@ in an external magnetic field of5 xx 10^(-4) T . What is the work done by the magnetic field in bringing it in its direction ? [ The magnetic moment = 20 A-m^(2)] |
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Answer» Solution :Work DONE by the MAGNETIC field, ` W = - MB( cos theta_1 - cos theta_2)` Here `theta_1 = 60^@ and theta_2 = 0` `:. W = -20 xx 5 xx 10^(-4) [ cos 60^(@) - cos 0]` `=-10^(-2)[(1)/(2)-1] = 5 xx 10^(-3) J`. |
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