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Question -

A closely wound solenoidof 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 toturn in a horizontal plane.

(a) What is the magneticmoment associated with the solenoid?

(b) What is the force andtorque on the solenoid if a uniform horizontal magnetic field of 7.5 × 10−2 Tis set up at an angle of 30º with the axis of the solenoid? 



Answer -

Number of turnson the solenoid, n = 2000

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

Current in thesolenoid, I = 4 A

(a)The magnetic momentalong 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 themagnetic field and the axis of the solenoid, θ = 30°

Torque,
Since the magnetic field is uniform, the force on the solenoid iszero. The torque on the solenoid is 

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