Answer –
It is given that the number of turns in the solenoid, n = 800
Area of cross section is given by, A = 2.5×10-4 m2
And the current in the solenoid, I = 3.0 A
Because a magnetic field develops along its axis, or length, a current-carrying solenoid functions like a bar magnet. The magnetic moment of a current-carrying solenoid is computed as follows:
M = n I A
\[M=800\times 3\times 2.5\times {{10}^{-4}}\]
\[M=0.6J{{T}^{-1}}\]