A closely wound solenoid of 1000 turns and area of cross-section `2 xx 10^(-4) m^(2)` carries a current of 1A. It is placed in a horizontal plane with its axis making an angle of `30^(@)` with the direction of uniform magnetic field of `0.16 T`. Calculate the torque acting on the solenoid.

Explanation

Given : `A = 2 xx 10^(-4) m^(2),I = 1A, theta = 30^(@)`
`B = 0.16 T, n = 1000`
As we know,
`M = nIA`
`= 1000 xx 1 xx 2 xx 10^(-4)`
`M = 0.2 Am^(2)`
Now, `tau = MB sin theta`
`= 0.2 xx 0.16 xx sin 30^(@)`
`= 0.2 xx 0.16 xx 1/2`
`tau = 0.016 Nm`.