InterviewSolution
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InterviewSolution
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If the solenoid is free to turn about the vertical direction and a uniform horizontal magnetic field of 0.25 T is applied, what is the magnitude of torque on the solenoid when its axis makes an angle of `30^(@)` with the direction of applied field ? Here `M = 0.6 J T^(-1)` |
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Answer» Here `M = 0.6 J T^(-1)` `B = 0.25 T r = ? q = 30^(@)` As `r = m B sin theta :. r = 0.6 xx 0.25 sin 30^(@) = 0.075 N.m`. |
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