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If the solenoid in Exercise 5.5 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 ? |
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Answer» Solution :We have MAGNETIC moment m = 0.6 A `m^2` . Further more B=0.25 T and `theta = 30^@` . `therefore ` Magnetide of torque `tau= m B sin theta - 0.6 xx 0.25 xx sin 30^@ = 0.075 N m= 7.5xx10^(-2) J`. |
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