Question

A closely wound solenoid of 2000 turns and area of cross-section $1.6\times {10}^{-4}{m}^2$, 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 $7.5\times {10}^{-2}T$ is set up at an angle of ${30}^o$ with the axis of the solenoid?

Answer 1

$a)Given,$

Number of turns of coil, $N=2000$

Area of cross-section of solenoid, $A=1.6\times {10}^{-4}{m}^2$

Current passing through the coil,$I=4A$

Magnetic moment, $M=NIA$

$=2000\times 4\times 1.6\times {10}^{-4}$

$=1.28A{m}^2$

The direction of $\overrightarrow{M}$ is along the axis of the solenoid in the direction in the related to the sense of current via the right-handed screw rule.

$b)$ Uniform magnetic field applies, $B=7.5\times {10}^{-2}T$

The angle between the axis of the solenoid and magnetic field $\theta ={30}^0$

Since the magnetic field is uniform on the solenoid, the force acting on the solenoid is $0$.

Torque is given by,

$\tau =MBsin\theta$

$=1.28\times 7.5\times {10}^{-2}sin{30}^0$

$=1.28\times 7.5\times {10}^{-2}\times \frac{1}{2}J$

$=0.048J$
The direction of the torque is such that the solenoid tends to align the axis of the solenoid (magnetic moment vector) along the direction of the magnetic field $\overrightarrow{B}$