A closely wound solenoid of 2000 turns and area of cross section `1.6xx10^-4m^2`, carrying a current of `4amp.` 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 are the force and torque on the solenoid if a uniform horizontal magnetic field of `7*5xx10^-2T` is set up at an angle of `30^@` with the axis of the solenoid?

Explanation

(a) The magnetic moment of the solenoid of N turns, area of cross section A and carrying current I is given by:
M = NIA
Here, N=2,000 and I = 4 A, `A=1.6 xx 10^(-4) m^2`
` rArr M = 2,000 xx 4.0 xx 1.6 xx 10^(-4) = 1.28 Am^2`
Direction of the magnetic moment, as per the sense of current, is along the axis of the solenoid according to the right-hand rule.
(b) Net force on the solenoid will be zero as the field is uniform. Torque acting on the solenoid is given by
` tau = MB sin theta `
` = 2.28 xx 7.5 xx 10^(-2) xx sin 30^@`
` = 1.28 xx 7.5 xx 10^(-3) xx 1/2 = 4.8 xx 10^(-2) Nm`
It will tend to align the magnetic moment vector along the direction of the magnetic field.