NCERT Class XII Chapter
Magnetism and Matter
Questions With Solutions
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Question : 6
Total: 25
If the solenoid in the previous question 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?
Solution:
The solenoid behaves as a bar magnet.
so, torque, τ = MB sinθ
τ = 0.6 × 0.25 × sin 30° = 0.075 N m.c
so, torque, τ = MB sinθ
τ = 0.6 × 0.25 × sin 30° = 0.075 N m.c
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