To manufacture a solenoid of length $1 \mathrm{~m}$ and inductance $1 \mathrm{mH}$, the length of thin wire…
To manufacture a solenoid of length $1 \mathrm{~m}$ and inductance $1 \mathrm{mH}$, the length of thin wire required is
(cross - sectional diameter of a solenoid is considerably less than the length)
$0.10 \mathrm{~m}$
$0.10 \mathrm{~km}$
$1 \mathrm{~km}$
$10 \mathrm{~km}$
Solution
Inductance of solenoid, $\mathrm{L}=\frac{\mu_0 \mathrm{~N}^2 \mathrm{~A}}{l}$ where, $l=$ length of solenoid.
$\mathrm{A}=\pi \mathrm{r}^2=$ area of solenoid
Let ' $x$ ' be length of wire required.
$\therefore \quad \mathrm{x}=$ circumference of solenoid $\times$ no. of turns $=2 \pi \mathrm{rN}$.
$\therefore \quad \mathrm{N}=\frac{\mathrm{x}}{2 \pi \mathrm{r}}$
Substituting in equation (i),
$\begin{aligned}
\mathrm{L} & =\frac{\mu_0\left(\frac{\mathrm{x}^2}{4 \pi^2 \mathrm{r}^2}\right) \times \pi \mathrm{r}^2}{l} \\
\therefore \quad \mathrm{L} & =\frac{\mu_0 \mathrm{x}^2}{4 \pi l} \\
\therefore \quad \mathrm{x}^2 & =\frac{4 \pi \mathrm{L} l}{\mu_0} \\
\therefore \quad \mathrm{x} & =\sqrt{\frac{4 \pi \mathrm{L} l}{\mu_0}}
\end{aligned}$
Substituting the values,
$\begin{aligned}
& x=\sqrt{\frac{4 \times \pi \times 10^{-3} \times 1}{4 \pi \times 10^{-7}}} \\
& x=\sqrt{10^4} \mathrm{~m} \\
& x=0.10 \mathrm{~km}
\end{aligned}$