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Question

A closely wound solenoid of 800 turns and area of cross section2.5 × 10–4 m2 carries a current of 3.0 A. Explain the sense in which the solenoid acts like a bar magnet. What is its associated magnetic moment?

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Solution

Given: number of turns of the solenoid is 800, the cross-sectional area of the solenoid is 2.5× 10 4 m 2 and current flowing through the solenoid is 3.0A.

The magnetic dipole moment of a solenoid is given as,

M=NIA

Substitute the values in the above equation, we get,

M=800×3×2.5× 10 4 =0.6 JT -1

Thus, the magnetic dipole moment is 0.6 JT -1 .

Since magnetic field is developed along the axis of the solenoid, thus the solenoid acts as a bar magnet.


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