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, Areas of cross section of the solenoid `A=1.6xx10^(-4)m^(2)`
Crurent the solenoid I=4A
(i) The magnetic moment along the axis of the solenoid is calculated as :
`M=nAI=2000xx1.6xx10^(-4)xx4=1.24Am^(2)`
(ii) Magnetic field `B=7.5xx10^(-2)T`
Angle between the magnetic fiedl and the axis of the solenoid `theta=30^(@)`
Torque, `tau=Mbsin theta = 1.28xx7.5xx10^(-2)sin 30^(@), =4.8xx10^(-2Nm`
Since the magnetic field is uniform, the force on the solenoid is zero. The torque on the solenoid is `4.8xx10^(-2)Nm^(5)`.