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?

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 3.0 A.

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.