The magnetic field $B$ inside a long solenoid carrying current $I$ with $n$ turns per unit length is given by the formula: $$B = \mu_{0} n I$$
The magnetic flux $\phi$ through a single turn of the solenoid is the product of the magnetic field and the cross-sectional area $A$. Given the radius $r$, the area is $A = \pi r^{2}$. $$\phi = B \cdot A = (\mu_{0} n I)(\pi r^{2})$$
The total number of turns $N$ in a solenoid of length $l$ is $N = n \cdot l$. The total flux linkage $\Phi$ is: $$\Phi = N \cdot \phi = (n l) \cdot (\mu_{0} n I \pi r^{2}) = \mu_{0} n^{2} I \pi r^{2} l$$
The self-inductance $L$ is defined by the relation $\Phi = L I$. Therefore: $$L = \frac{\Phi}{I} = \frac{\mu_{0} n^{2} I \pi r^{2} l}{I} = \mu_{0} \pi n^{2} r^{2} l$$
Final Answer: \mu_{0}\pi n^{2}r^{2}l
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