1.

If the solenoid in is free to turn about the vertical direction and a uniform horizontal magnetic field of 0.25T 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 ? Given magnetic moment 0.6 JT−1 .

Answer»

SOLUTION :B = 0.25 T , `theta =30^@`
`taumB sin theta=0.6xx0.25xxsin30=0.6xx0.25xx1/2=0.075Nm ` (or J)


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