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A $50 \mathrm{~cm}$ long solenoid has winding of 400 turns. What current must pass through it to produce a magnetic field of induction $4 \pi \times 10^{-3} \mathrm{~T}$ at the centre?
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Verified Answer
The correct answer is:
12.5A
For a solenoid,
$B=\mu_0 n I \Rightarrow n=\frac{N}{L}$
Here,
$\begin{aligned}
& B=4 \pi \times 10^{-3} \mathrm{~T} \\
& n=\frac{400}{50 \times 10^{-2}}=\frac{400 \times 100}{50}=800 \text { turns } / \mathrm{m}
\end{aligned}$
$\begin{aligned}
& \text { Since, } B=\mu_0 n I \\
& 4 \pi \times 10^{-3}=4 \pi \times 10^{-7} \times 800 \times I \\
& \Rightarrow \quad I=\frac{4 \pi \times 10^{-3}}{4 \pi \times 10^{-7} \times 800} \Rightarrow I=\frac{10000}{800}=12.5 \mathrm{~A}
\end{aligned}$
$B=\mu_0 n I \Rightarrow n=\frac{N}{L}$
Here,
$\begin{aligned}
& B=4 \pi \times 10^{-3} \mathrm{~T} \\
& n=\frac{400}{50 \times 10^{-2}}=\frac{400 \times 100}{50}=800 \text { turns } / \mathrm{m}
\end{aligned}$
$\begin{aligned}
& \text { Since, } B=\mu_0 n I \\
& 4 \pi \times 10^{-3}=4 \pi \times 10^{-7} \times 800 \times I \\
& \Rightarrow \quad I=\frac{4 \pi \times 10^{-3}}{4 \pi \times 10^{-7} \times 800} \Rightarrow I=\frac{10000}{800}=12.5 \mathrm{~A}
\end{aligned}$
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