Inductance of solenoid ' $L$ ' having diameter ' $d$ '. Let $n$ be the number of turns per unit length. The…
Inductance of solenoid ' $L$ ' having diameter ' $d$ '. Let $n$ be the number of turns per unit length. The inductance per length near the middle of a solenoid is (Asume that current passes through the turns, $\mu_0=$ Permeability of vacuum)
$\frac{\mu_0 \pi n d^2}{4}$
$\frac{\mu_0 \pi n^2 d}{2}$
$\frac{\mu_0 \pi n^2 d^2}{2}$
$\frac{\mu_0 \pi n^2 d^2}{4}$
Solution
Concept: Using the definition of the flux associated through the surface we can obtain the inductance per unit length near the center of the solenoid.
We define selfinductance $\mathrm{L}$ as:
$\phi=\mathrm{Li}$
where, $\phi$ is the associated flux and $\mathrm{i}$ is the current through the solenoid.
$\phi=\mathrm{B} \cdot \mathrm{A}$
where, $B=\left(\mu_0\right.$ in $), A=2 \pi \frac{d^2}{4}$ is the area.
$\begin{aligned} & \therefore \mathrm{Li}=\left(\mu_0 \mathrm{in}\right)\left(\mathrm{n} \pi \frac{\mathrm{d}^2}{4}\right) \\ & \therefore \mathrm{L}=\frac{\mu_0 \pi \mathrm{n}^2 \mathrm{~d}^2}{4}\end{aligned}$