Question
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,
= 800 x 3.0 x 2.5 x 10–4
= 0.6 JT–1
Torque acting on the solenoid,
= 0.150 x 0.5 = 7.5 x 10–2 J.
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,
= 800 x 3.0 x 2.5 x 10–4
= 0.6 JT–1
Torque acting on the solenoid,
= 0.150 x 0.5 = 7.5 x 10–2 J.