$A$ closely wound solenoid of $800$ turns and area of cross-section $2.5 \times 10^{-4} \; m^2$ 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?

(N/A) Number of turns in the solenoid,$n = 800$.
Area of cross-section,$A = 2.5 \times 10^{-4} \; m^2$.
Current in the solenoid,$I = 3.0 \; A$.
$A$ current-carrying solenoid behaves as a bar magnet because a magnetic field is produced along its axis,similar to the magnetic field lines of a bar magnet.
The magnetic moment $(M)$ associated with the solenoid is given by the formula: $M = nIA$.
Substituting the values: $M = 800 \times 3.0 \times 2.5 \times 10^{-4}$.
$M = 2400 \times 2.5 \times 10^{-4} = 6000 \times 10^{-4} = 0.6 \; J \cdot T^{-1}$.