If the solenoid in the above question is free to turn about the vertical direction, and a uniform horizontal magnetic field of `0*25T` is applied, what is the magnitude of the torque on the solenoid when its axis makes an angle of `30^@` with the direction of the applied field?

Options

A. 0.075Nm
B. 0.080Nm
C. 0.081Nm
D. 0.091Nm

Answer

C

Explanation

`Here, M=0.65JT^(-1), B=0.25T, theta=30^(@)`
`therefore tau=mB sin theta=0.65xx0.25xxsin 30^(@)`
`=0.65xx0.25xx(1)/(2)=0.08125Nm`
=0.081 Nm