A closely wound solenoid of 2000 turns and area of cross-section 1.6×10^-4 m², carrying 4.0 A current, is suspended through its centre, thereby 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 the horizontal magnetic field of 7.5 × 10^–2 T is set up at an angle of 30° with the axis of the solenoid?

Answer –

We are given,

Number of turns on the solenoid is given by n = 2000

Area of cross-section of the solenoid is given by A = 1.6×10-4 m²

Current in the solenoid is I = 4.0 A

• The calculation of magnetic moment along the axis of the solenoid is carried out by:

M = nAI, therefore –

M = 2000 × 1.6×10-4 × 4.0

M = 1.28 Am²

• Magnetic field is given by B = 7.5 × 10^–2 T

The angle made with the axis of solenoid is given by –

\[\theta ={{30}^{\circ }}\]

\[\therefore T=MB\sin \theta \]

\[T=1.28\times 7.5\times {{10}^{-2}}\times \sin {{30}^{\circ }}\]

So, Torque (T) = 0.048 J

The force on the solenoid is 0 because the magnetic field is homogenous.The torque on the solenoid is 0.048 J.