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A long solenoid has 20 turns $\mathrm{cm}^{-1}$. The current necessary to produce a magnetic field of $20 \mathrm{mT}$ inside the solenoid is approximately
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2644 Upvotes
Verified Answer
The correct answer is:
$8 \mathrm{~A}$
Given, $n=20 \mathrm{~cm}^{-1}=2000 \mathrm{~m}^{-1}$
$$
B=20 \mathrm{mT}=20 \times 10^{-3} \mathrm{~T}
$$
$$
\begin{aligned}
\text { Using, } B &=\mu_{0} n I \\
20 \times 10^{-3} &=4 \pi \times 10^{-7} \times 2000 \times I \\
\Rightarrow \quad I &=\frac{20 \times 10^{-3}}{4 \pi \times 10^{-7} \times 2000}=8 \mathrm{~A}
\end{aligned}
$$
$$
B=20 \mathrm{mT}=20 \times 10^{-3} \mathrm{~T}
$$
$$
\begin{aligned}
\text { Using, } B &=\mu_{0} n I \\
20 \times 10^{-3} &=4 \pi \times 10^{-7} \times 2000 \times I \\
\Rightarrow \quad I &=\frac{20 \times 10^{-3}}{4 \pi \times 10^{-7} \times 2000}=8 \mathrm{~A}
\end{aligned}
$$
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