Inductance per unit length near the middle of a long solenoid is $\left(\mu_0=\right.$ permeability of free…

Inductance per unit length near the middle of a long solenoid is $\left(\mu_0=\right.$ permeability of free space, $\mathrm{n}=$ number of turns per unit length, $\mathrm{d}=$ the diameter of the solenoid)
  1. $\mu_0 \pi\left(\frac{\mathrm{nd}}{2}\right)^2$
  2. $4 \mu_0 \pi\left(\frac{\mathrm{nd}}{2}\right)$
  3. $\left(\frac{\mu_0 \pi \mathrm{nd}}{2}\right)$
  4. $\frac{4 \mu_0 \pi}{n^2 d^2}$

Solution

The inductance long solenoid is $\begin{aligned} & \mathrm{L}=\frac{\mu_0 \mathrm{~N}^2 \mathrm{~A}}{l} \\ & \mathrm{~L}=\mu_0\left(\frac{\mathrm{N}}{l}\right)^2 \times \pi \times \frac{\mathrm{d}^2}{4} \end{aligned}$ $\therefore \quad$ The inductance per unit length near the middle of a long solenoid is: $\frac{\mathrm{L}}{l}=\mu_0 \pi\left(\frac{\mathrm{nd}}{2}\right)^2 \quad \ldots .\left(\because \frac{\mathrm{N}}{l}=\mathrm{n}\right)$

Asked in: MHT CET 2023 (12 May Shift 2)

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