Q129 / 150 · MHT CET 2021 (24 Sep Shift 1) · Physics
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$3 \times 10^{-3} \mathrm{H}$
- Correct answerB$1 \times 10^{-3} \mathrm{H}$
- C$2 \times 10^{-3} \mathrm{H}$
- D$4 \times 10^{-3} \mathrm{H}$
Solution
$$
\begin{aligned}
& \ell=31.4 \mathrm{~cm}=0.314 \mathrm{~m} \\
& \mathrm{~A}=10^{-3} \mathrm{~m}^2, \mathrm{~N}=500
\end{aligned}
$$
Self-inductance, $L=\mu_0 n^2 \ell A$
$$
\begin{aligned}
& =\frac{\mu_0 \mathrm{~N}^2 \mathrm{~A}}{\ell}, \quad \text { where } \mathrm{n}=\frac{\mathrm{N}}{\ell} \\
& =\frac{4 \pi \times 10^{-7} \times(500)^2 \times 10^{-3}}{0.314} \\
& =\frac{4 \times 3.14 \times 25 \times 10^{-6}}{0.314} \\
& =10^{-3} \mathrm{H}
\end{aligned}
$$
Hence, the correct answer is option: 'B'.