Question
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
The direction of 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, = 30°
Since the magnetic field is uniform on the solenoid, force acting on the solenoid is 0.
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
The direction of 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, = 30°
Since the magnetic field is uniform on the solenoid, force acting on the solenoid is 0.
Torque is given by,
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 .
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 .