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(D) The magnetic field inside a long solenoid is given by $B = \mu_0 n i$,where $n$ is the number of turns per unit length and $i$ is the current.The

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$\mu_0 n^2 \pi R^2$

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(D) The magnetic field inside a long solenoid is given by $B = \mu_0 n i$,where $n$ is the number of turns per unit length and $i$ is the current.The magnetic flux $\phi$ through each turn of the solenoid is $\phi = B A = (\mu_0 n i)(\pi R^2)$.For a solenoid of length $l$,the total number of turns $N$ is $N = n l$.The total magnetic flux linkage is $N \phi = (n l)(\mu_0 n i \pi R^2) = \mu_0 n^2 i \pi R^2 l$.The self-inductance $L$ is defined as $L = \frac{N \phi}{i} = \frac{\mu_0 n^2 i \pi R^2 l}{i} = \mu_0 n^2 \pi R^2 l$.Therefore,the self-inductance per unit length is $\frac{L}{l} = \mu_0 n^2 \pi R^2$.

$A$ solenoid of radius $R$ has $n$ turns per unit length. The self-inductance of the solenoid per unit length is:

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