Question

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.

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 30° with the axis of the solenoid?

Answer 1

Magnetic field, B = 7.5 × 10⁻² T

Angle between the magnetic field and the axis of the solenoid, θ = 30°

`"Torque", tau = "MB  sin"  theta`

`= 1.28 xx 7.5 xx 10^-2  "sin"  30°`

`= 4.8 xx 10^-2  "Nm"`

Since the magnetic field is uniform, the force on the solenoid is zero. The torque on the solenoid is `4.8 xx 10^-2  "Nm"`