5.8 A closely wound solenoid of 2000 turns and area of cross-section 1.6 × 10–4 m2, carrying a current of 4.0 A, is suspended through its center 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 × 10–2 T is set up at an angle of 30° with the axis of the solenoid?

Number of turn on the solenoid, n=2000
Area of cross-section of the solenoid,A=1.6×10-4m2
Current In the solenoid, I = 4A
(a) The magnetic field moment along the axis of the solenoid is calculated as:M=n A I=2000×1.6×10-4×4=1.28 Am2
(b) Magnetic field,
B=7.5×10-2 T
Angle between the magnetic field and the axis of the solenoid,θ=30°
Torque,τ=MB sin θ=1.28×7.5×102sin30=4.8×102Nm
Since the magnetic field is uniform, the force on the solenoid is zero. The torque on the solenoid is4.8×10-2 Nm.