Number of turns in the solenoid, \(n=800\)
Area of cross-section, \(A=2.5 \times 10^{-4} \mathrm{~m}^{2}\)
Current in the solenoid, \(I=3.0 \mathrm{~A}\)
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:
\(M=n I \mathrm{~A}\)
\(=800 \times 3 \times 2.5 \times 10^{-4}\)
\(=0.6 \mathrm{~J} \mathrm{~T}^{-1}\)
