Class 12 Physics
CBSE and UP Board
Magnetism and Matter
Chapter-5 Exercise 5.6
NCERT Solutions for Class 12 Physics Chapter 5 Question 6
If the solenoid in Exercise 5.5 is free to turn about the vertical direction and a uniform horizontal magnetic field of 0.25 T is applied, what is the magnitude of torque on the solenoid when its axis makes an angle of 30° with the direction of applied field?
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Magnetic field strength, B = 0.25 T
Magnetic moment, M = 0.6 T⁻¹
The angle 0, between the axis of the solenoid and the direction of the applied field is 30°.
Therefore, the torque acting on the solenoid is given as:
τ = MB sin O
= 0.6 x 0.25 sin 30°
= 7.5 x 10⁻2 J