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.
(a) What is the magnetic moment associated with the solenoid?
(b) 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

(a) Magnetic moment of current carrying solenoid,
`m_(s) = NIA`
`therefore m_(s) = (2000) (4) (1.6 xx 10^(-4) )`
`=1.28 Am^(2)`
(b) (i) Resultant force exerted on a magnet in a uniform exerted magnetic field is,
`overset(to)(F) = -q_(m) overset(to)(B) + q_(m) overset(to) (B) = overset(to) (0)`
(Where, `q_(m)=` pole strength of either pole of a magnet)
(ii) Torque exerted on above solenoid,
`tau = m_(s) B sin theta`
`= (1.28) (7.5 xx 10^(-2) ) (sin 30^(@) )`
`= (1.28 ) (0.075 ) (0.5)`
`= 0.048` Nm