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

Number of turns on the solenoid, n=2000
Area of cross-section of the solenoid, A=`1.6xx10^(-4)m^(2)`
Current in the solenoid, I=4A
(a) The magnetic moment along the axis of the solenoid is calculated as:
M=nAI
`=2000xx1.6xx10^(-4)xx4`
`=1.28Am^(2)`
(b) Mangetic field `B=7.5xx10^(-2)T`
angle between the magnetic field and the axis of the solenoid, `theta=30^(@)`
Torque, r=`MB sin theta`
`=1.28xx7.5xx10^(-7)sin30^(@)`
`=4.8xx10^(-2)Nm`
Since the magnetic field is uniform, the force on the solenoid is zero, the torque on the solenoid is `4.8xx10^(-2)Nm`.