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Magnetism And Matter

Question
CBSEENPH12037720

A closely wound solenoid of 2000 turns and area of cross-section 1.6 x 10–4 m2, 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 x 10–2 T is set up at an angle of 30° with the axis of the solenoid?

Solution
(a) Given, 
Number of turns of coil, N = 2000
Area of cross-section of solenoid, A= 1.6 x 10–4 m2
Current passing through the coil, I = 4 A

Magnetic moment , M = NIA

                                  = 2000 × 4 × 1.6 × 10-4      =1.28 Am2 
The direction of  M is along the axis of the solenoid in the direction related to the sense of current via the right handed screw rule. 

(b) Uniform magnetic field applies, B = 7.5 x 10–2 T 
Angle between the axis of the solenoid and magnetic field, θ = 30°
Since the magnetic field is uniform on the solenoid, force acting on the solenoid is 0.

Torque is given by, 

 τ = MB sin θ    = 1.28 × 7.5 × 10-2 sin 30°   =1.28 ×7.5 ×10-2×12J    = 0.048 J. 
 
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 magnetic field B. 

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