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# A closely wound solenoid of 2000 turns and area of cross-section

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.

(a) What is the magnetic moment associated with the solenoid?

(b) 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 30º with the axis of the solenoid?

Solution:

Number of turns on the solenoid, n = 2000

Area of cross-section of the solenoid, A = 1.6 × 10−4 m2

Current in the solenoid, I = 4 A

(a)The magnetic moment along the axis of the solenoid is calculated as:

M = nAI

= 2000 × 1.6 × 10−4 × 4

= 1.28 Am2

(b)Magnetic field, B = 7.5 × 10−2 T

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

Torque, $\tau=M B \sin \theta$

$=1.28 \times 7.5 \times 10^{-2} \sin 30^{\circ}$

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

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