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Magnetism And Matter

Question
CBSEENPH12037718

If the solenoid in Question 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?

Solution
Given,
Number of turns, N = 800
Current passing through solenoid, I = 3A
Area of cross-section, A = 2.5 x 10
–4 m2
Magnetic field, B = 0.25 T
Angle between the axis and B, θ = 30° 

Magnetic moment, M = N.I.A.
                                 = 800 x 3.0 x 2.5 x 10–4
                                 = 0.6 JT–1 

Torque acting on the solenoid, τ = MB sin θ
                                                  = 0.6 × 0.25 × sin 30°
                                                = 0.150 x 0.5 = 7.5 x 10–2 J.