Class 12 Physics
CBSE and UP Board
Magnetism and Matter
Chapter-5 Exercise 5.8
NCERT Solutions for Class 12 Physics Chapter 5 Question 8
A closely wound solenoid of 2000 turns and area of cross-section 1.6 × 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 × 10⁻² T is set up at an angle of 30º with the axis of the solenoid?
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Number of turns on the solenoid, n = 2000
Area of cross-section of the solenoid, A = 1.6 x 10-4m2
Current in the solenoid, I = 4 A
Ans (a).
The magnetic moment along the axis of the solenoid is calculated as:
M = nAI = 2000 x 1.6 x 10⁻4 x 4 = 1.28 Am2
Ans (b).
Magnetic field, B = 7.5 x 10-2 T
Angle between the magnetic field and the axis of the solenoid, 0 = 30°
Torque, τ = MBsinB
= 1.28 x 7.5 x10⁻2 sin 30°
= 4.8 x 10⁻2 Nm
Since the magnetic field is uniform, the force on the solenoid is zero. The torque on the solenoid is 4.8 x 10⁻2 Nm.