A closely wound solenoid of 2000 turns and area of cross-section `1.6 xx 10^(-4)m^(2)`, carrying a current of 4.0 A, is suspended through its centre allowing it to turn in a horizontal plane.
What is the force and torque on the solenoid if a uniform horizontal magnetic field of `7.5 xx 10^(-2)T` is set up at an angle of `30^(@)` with the axis of the solenoid ?

Explanation

Net force on the solenoid = 0
Torque, `r = m B sin theta = 1.28 xx 7.5 xx 10^(-2) sin 30^(@)`
`= 1.28 xx 7.5 xx 10^(-2)xx (1)/(2)`
`r = 4.8 xx 10^(-2) Nm`.