How many meters of a thin wire are required to manufacture a solenoid of length l = 100 cm and inductance L = 1 mH,  if the solenoid's cross-sectional diameter is considerably less than its length?

Magnetic induction at centre of solenoid is

$\mathrm{B}=\frac{{\mathrm{\mu }}_0\mathrm{N}}{{\mathrm{l}}_0}\mathrm{I}$

Then flux linked with whole solenoid is

$\mathrm{\varphi }=\mathrm{NBA}=\frac{{\mathrm{\mu }}_0{\mathrm{N}}^2{\mathrm{\pi r}}^2\mathrm{I}}{{\mathrm{l}}_0}$                   …(i)

Also, $\mathrm{\varphi }=\mathrm{LI}$                                    ….(ii)

$\therefore \mathrm{L}=\frac{{\mathrm{\mu }}_0{\mathrm{N}}^2{\mathrm{\pi r}}^2}{{\mathrm{l}}_0}$

$\text{ or }\mathrm{L}={\mathrm{\mu }}_0\frac{{\mathrm{N}}^2\cdot 4\mathrm{\pi }.{\mathrm{\pi r}}^2}{4{\mathrm{\pi l}}_0}={\mathrm{\mu }}_0\frac{{\mathrm{N}}^2\cdot 4{\mathrm{\pi }}^2\cdot {\mathrm{r}}^2}{4{\mathrm{\pi l}}_0}$

$\Rightarrow \mathrm{L}={\mathrm{\mu }}_0\cdot \frac{(\mathrm{N}\cdot 2\mathrm{\pi r}{)}^2}{4{\mathrm{\pi l}}_0}$

Now, N. $2\mathrm{\pi r}=\mathrm{l}=$ length of wire used in solenoid

$\therefore \mathrm{L}=\frac{{\mathrm{\mu }}_0\cdot {\mathrm{l}}^2}{4{\mathrm{\pi l}}_0}$

$\mathrm{l}=\sqrt{\frac{4{\mathrm{\pi l}}_0\mathrm{L}}{{\mathrm{\mu }}_0}}$

Then, l = 0.10 km