MHT CET · Physics · Electromagnetic Induction
The self-inductance of solenoid of length \(31.4 \mathrm{~cm}\), area of cross section \(10^{-3} \mathrm{~m}^2\) having total number of turns 500 will be nearly \(\left[\mu_0=4 \pi \times 10^{-7}\right.\) SI unit \(]\)
- A \(3 \times 10^{-3} \mathrm{H}\)
- B \(1 \times 10^{-3} \mathrm{H}\)
- C \(2 \times 10^{-3} \mathrm{H}\)
- D \(4 \times 10^{-3} \mathrm{H}\)
Answer & Solution
Correct Answer
(B) \(1 \times 10^{-3} \mathrm{H}\)
Step-by-step Solution
Detailed explanation
\(
\begin{aligned}
& \ell=31.4 \mathrm{~cm}=0.314 \mathrm{~m} \\
& \mathrm{~A}=10^{-3} \mathrm{~m}^2, \mathrm{~N}=500
\end{aligned}
\)
Self-inductance, \(L=\mu_0 n^2 \ell A\)
\(
\begin{aligned}
& =\frac{\mu_0 \mathrm{~N}^2 \mathrm{~A}}{\ell}, \quad \text { where } \mathrm{n}=\frac{\mathrm{N}}{\ell} \\
& =\frac{4 \pi \times 10^{-7} \times(500)^2 \times 10^{-3}}{0.314} \\
& =\frac{4 \times 3.14 \times 25 \times 10^{-6}}{0.314} \\
& =10^{-3} \mathrm{H}
\end{aligned}
\)
\begin{aligned}
& \ell=31.4 \mathrm{~cm}=0.314 \mathrm{~m} \\
& \mathrm{~A}=10^{-3} \mathrm{~m}^2, \mathrm{~N}=500
\end{aligned}
\)
Self-inductance, \(L=\mu_0 n^2 \ell A\)
\(
\begin{aligned}
& =\frac{\mu_0 \mathrm{~N}^2 \mathrm{~A}}{\ell}, \quad \text { where } \mathrm{n}=\frac{\mathrm{N}}{\ell} \\
& =\frac{4 \pi \times 10^{-7} \times(500)^2 \times 10^{-3}}{0.314} \\
& =\frac{4 \times 3.14 \times 25 \times 10^{-6}}{0.314} \\
& =10^{-3} \mathrm{H}
\end{aligned}
\)
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