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°

${\displaystyle \text{Torque},\tau =\text{MB sin} \theta }$

${\displaystyle =1.28\times 7.5\times {10}^{-2} \text{sin} 30°}$

${\displaystyle =4.8\times {10}^{-2} \text{Nm}}$

Since the magnetic field is uniform, the force on the solenoid is zero. The torque on the solenoid is ${\displaystyle 4.8\times {10}^{-2} \text{Nm}}$