If the solenoid in Exercise 5.5 is free to tum about the vertical direction and a uniform hortzontal 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?

Explanation

Above solenoid acts like a bar magnet with magnetic dipole moment `m_(s) = 0.6 Am^(2)`. Now, torque exerted on it by uniform external magnetic field is,
`tau= m_(s) B sin theta`
`= (0.6) (0.25) sin 30^@`
`=(0.6) (0.25) (0.5)`
`tau= 0.075` Nm