MHT CET · Physics · Magnetic Effects of Current
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)
- A \(\mu_0 \pi\left(\frac{\mathrm{nd}}{2}\right)^2\)
- B \(4 \mu_0 \pi\left(\frac{\mathrm{nd}}{2}\right)\)
- C \(\left(\frac{\mu_0 \pi \mathrm{nd}}{2}\right)\)
- D \(\frac{4 \mu_0 \pi}{n^2 d^2}\)
Answer & Solution
Correct Answer
(A) \(\mu_0 \pi\left(\frac{\mathrm{nd}}{2}\right)^2\)
Step-by-step Solution
Detailed explanation
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:
\(
\frac{\mathrm{L}}{l}=\mu_0 \pi\left(\frac{\mathrm{nd}}{2}\right)^2 \quad \ldots .\left(\because \frac{\mathrm{N}}{l}=\mathrm{n}\right)
\)
\(
\begin{aligned}
& \mathrm{L}=\frac{\mu_0 \mathrm{~N}^2 \mathrm{~A}}{l} \\
& \mathrm{~L}=\mu_0\left(\frac{\mathrm{N}}{l}\right)^2 \times \pi \times \frac{\mathrm{d}^2}{4}
\end{aligned}
\)
\(\therefore \quad\) The inductance per unit length near the middle of a long solenoid is:
\(
\frac{\mathrm{L}}{l}=\mu_0 \pi\left(\frac{\mathrm{nd}}{2}\right)^2 \quad \ldots .\left(\because \frac{\mathrm{N}}{l}=\mathrm{n}\right)
\)
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