A closely wound solenoid of 2000 turns and are of cross - section `1.6xx10^(-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 `7.5xx10^(-2)T` is set up at an angle of `30^@` with the axis of the solenoid ?

Explanation

N = 200 , A `1.6xx10^(-4)m^2,I=4.0A`
a. `m=ANI = 1.6 xx10^(-4) xx2000xx4.0=1.28Am^2` , along the axis
b. `B = 7.5 xx10^(-2)T, theta = 30^@`
Net force = 0
`tau=mBsintheta=1.28xx7.5xx10^(-2)xxsin30=0.64xx7.5xx10^(-2)=4.800xx10^(-2)Nm`
By the action of this, the solenoid can come to the direction of external field.