MHT CET · Physics · Electromagnetic Induction
An air cored coil has a self inductance 0.1 H . A soft iron core of relative permeability 1000 is introduced and the number of turns is reduced \(\left(\frac{1}{10}\right)^{\text {th }}\). The value of self inductance is
- A 0.1 H
- B 1 mH
- C 1 H
- D 10 mH
Answer & Solution
Correct Answer
(C) 1 H
Step-by-step Solution
Detailed explanation
For air core coil, \(L=0.1=\frac{\mu_0 \mathrm{~N}^2 \mathrm{~A}}{l}\)
For soft iron core coil, \(\mathrm{L}^{\prime}=\frac{\mu_0 \mu_{\mathrm{r}}\left(\mathrm{N}^{\prime}\right)^2 \mathrm{~A}}{l}\)
\(\begin{aligned}
& \frac{L^{\prime}}{L}=\frac{1000 \times \mu_0 \times\left(\frac{N}{10}\right)^2 \mathrm{~A} \times l}{\mu_0 \mathrm{~N}_2 \mathrm{~A} \cdot l} \\
& \frac{\mathrm{~L}^{\prime}}{\mathrm{L}^{\prime}}=10 \\
& \mathrm{~L}^{\prime}=0.1 \times 10=1 \mathrm{H}
\end{aligned}\)
For soft iron core coil, \(\mathrm{L}^{\prime}=\frac{\mu_0 \mu_{\mathrm{r}}\left(\mathrm{N}^{\prime}\right)^2 \mathrm{~A}}{l}\)
\(\begin{aligned}
& \frac{L^{\prime}}{L}=\frac{1000 \times \mu_0 \times\left(\frac{N}{10}\right)^2 \mathrm{~A} \times l}{\mu_0 \mathrm{~N}_2 \mathrm{~A} \cdot l} \\
& \frac{\mathrm{~L}^{\prime}}{\mathrm{L}^{\prime}}=10 \\
& \mathrm{~L}^{\prime}=0.1 \times 10=1 \mathrm{H}
\end{aligned}\)
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