A solenoid is designed to generate a magnetic field over a large volume. Its dimensions are as follows: length $=2$ meters, radius $=0.1$ meter, number of turns $=1000$.
(Edge effects should be neglected.)

Calculate the self-inductance of the solenoid in Henrys.
inside it is

Suppose the solenoid carries a current $I$. The magnetic induction

$$
B=\mu_{0} n I=\mu_{0} N I / l,
$$

and the magnetic flux linkage is

$$
\psi=N B S=N \frac{\mu_{0} N I}{l} \cdot \pi r^{2}=\frac{I \mu_{0} N^{2} \pi r^{2}}{l} .
$$

Hence the self-inductance is

$$
\begin{aligned}
L=\frac{\psi}{I} &=\frac{\mu_{0} N^{2} \pi r^{2}}{l}=\frac{4 \pi \times 10^{-7} \times 1000^{2} \times \pi \times 0.1^{2}}{2} \\
&=1.97 \times 10^{-2} \mathrm{H} .
\end{aligned}
$$