(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.