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InterviewSolution
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A straight solenoid of length 50 cm has 1000 turns per metre and mean cross-sectional area of `2 xx 10^(-4) m^(2)` . It is placed with its axis of `30^(@)` with uniform magnetic field of 0.32 T . Find the torque acting on the solenoid when a current of 2 ampere is passed through it . |
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Answer» Torque of solenoid is given by `tau = MB sin theta` = (NiA) `B sin theta` `= 500 xx 2 xx 2 xx 10^(-4) xx 0.32 xx (1)/(2)` = 0.032 Nm |
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