Question:
A closely wound solenoid of 800 turns and area of cross section 2.5 × 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?
Solution:
Number of turns in the solenoid, n = 800
Area of cross-section, A = 2.5 × 10−4 m2
Current in the solenoid, I = 3.0 A
A current-carrying solenoid behaves as a bar magnet because a magnetic field develops along its axis, i.e., along its length.
The magnetic moment associated with the given current-carrying solenoid is calculated as:
M = n I A
= 800 × 3 × 2.5 × 10−4
= 0.6 J T−1