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?
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
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A short bar magnet of magnetic moment m = 0.32 J T-1 is placed in a uniform magnetic field of 0.15 T. If the bar is free to rotate in the plane of the field, which orientation would correspond to its ( a ) stable, and ( b) unstable equilibrium? What is the potential energy of the magnet in each case?
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Read More: Magnetism and Matter