Show the derivation to prove that solenoid acts as a bar magnet

$$\begin{gathered} \mathrm{Suppose} \mathrm{we} \mathrm{have} \mathrm{a} \mathrm{current} \mathrm{carrying} \mathrm{solenoid} \mathrm{of} \mathrm{length} 2\mathrm{l}. \\ \mathrm{and} \mathrm{radius} \mathrm{of} \mathrm{solenoid}=\mathrm{a} \\ \mathrm{Number} \mathrm{of} \mathrm{turns} \mathrm{per} \mathrm{unit} \mathrm{length}=\mathrm{n} \\ \mathrm{Current} \mathrm{passing} \mathrm{through} \mathrm{solenoid}=\mathrm{I} \\ \mathrm{Let} \mathrm{us} \mathrm{consider} \mathrm{a} \mathrm{point} \mathrm{P} \mathrm{outside} \mathrm{the} \mathrm{solenoid} \mathrm{at} \mathrm{distance} \mathrm{r} \mathrm{from} \mathrm{the} \mathrm{centre} \mathrm{of} \mathrm{solenoid} \text{'}\mathrm{O}\text{'}. \\ \mathrm{i}.\mathrm{e}. \mathrm{OP}=\mathrm{r} \\ \mathrm{Consider} \mathrm{a} \mathrm{small} \mathrm{element} \mathrm{of} \mathrm{thickness} \mathrm{dx} \mathrm{of} \mathrm{solenoid} \mathrm{at} \mathrm{distance} \mathrm{x} \mathrm{from} \mathrm{O}, \mathrm{then} \mathrm{number} \mathrm{of} \mathrm{turns} \mathrm{in} \mathrm{element}=\mathrm{ndx} \\ \mathrm{We} \mathrm{know} \mathrm{that} \mathrm{the} \mathrm{magnetic} \mathrm{field} \mathrm{due} \mathrm{to} \mathrm{n} \mathrm{turns} \mathrm{coil} \mathrm{at} \mathrm{axis} \mathrm{of} \mathrm{solenoid} \mathrm{is} \mathrm{given} \mathrm{by} \\ \mathrm{dB}=\frac{{\mathrm{\mu }}_0{\mathrm{Ia}}^2\left(\mathrm{ndx}\right)}{2\left[\right(\mathrm{r}-\mathrm{x}{)}^2+{\mathrm{a}}^2{]}^{3/2}} \\ \mathrm{If} \mathrm{r}>>\mathrm{a} \mathrm{and} \mathrm{r}>>\mathrm{x}, \\ \mathrm{then} \left[\right(\mathrm{r}-\mathrm{x}{)}^2+{\mathrm{a}}^2{]}^{3/2}={\mathrm{r}}^3 \\ \mathrm{Therefore}, \\ \mathrm{dB}=\frac{{\mathrm{\mu }}_0{\mathrm{Ia}}^2\left(\mathrm{ndx}\right)}{2{\mathrm{r}}^3} \\ \mathrm{The} \mathrm{total} \mathrm{magnetic} \mathrm{field} \mathrm{due} \mathrm{to} \mathrm{entire} \mathrm{solenoid} \mathrm{is} \mathrm{given} \mathrm{by} \\ \mathrm{B}=\frac{{\mathrm{\mu }}_0{\mathrm{Ia}}^2\left(\mathrm{n}\right)}{2{\mathrm{r}}^3}{\int }_{\mathrm{x}=-\mathrm{l}}^{\mathrm{x}=\mathrm{l}} \mathrm{dx} \\ =\frac{{\mathrm{\mu }}_0{\mathrm{Ia}}^2\left(\mathrm{n}\right)}{2{\mathrm{r}}^3}\times 2\mathrm{l}------\left(\mathrm{i}\right) \\ \mathrm{The} \mathrm{magnetic} \mathrm{moment} \mathrm{of} \mathrm{solenoid}, \mathrm{M}=\mathrm{NIA}=(\mathrm{n}\times 2\mathrm{l})\times \mathrm{I}\times \left({\mathrm{\pi r}}^2\right) \\ \mathrm{Or} (\mathrm{n}\times 2\mathrm{l})\times \mathrm{I}=\frac{\mathrm{M}}{\left({\mathrm{\pi r}}^2\right)} \\ \mathrm{Putting} \mathrm{it} \mathrm{in} \left(\mathrm{i}\right), \mathrm{we} \mathrm{get} \\ \mathrm{B}=\frac{{\mathrm{\mu }}_{0}2\mathrm{M}}{4{\mathrm{\pi r}}^3} \mathrm{or} \frac{{\mathrm{\mu }}_0}{4\mathrm{\pi }}\times \frac{2\mathrm{M}}{{\mathrm{r}}^3} \end{gathered}$$
Thus, it is clear from the above expression that magnetic moment of a bar magnet is equal to the magnetic moment of an equivalent solenoid that produce the same magnetic field or we can say that a solenoid acts as a bar magnet.