Question

A closely wound solenoid of 800 turns and area of cross section $2.5\times {10}^{-4}{m}^2$ carries a current of $3.0A$. Explain the sense in which the solenoid acts like a bar magnet. What is its associated magnetic moment?

Answer 1

Given: $n=800,$

$A=2.5\times {10}^{-4}{m}^2,$

$I=3.0A$

A magnetic field develops along the axis of the solenoid. Therefore current-carrying solenoid acts like a bar magnet.

$m=nIA=800\times 3\times 2.5\times {10}^{-4}J{T}^{-1}$

$=0.6J{T}^{-1}$ along the axis of the solenoid.