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A solenoid of length $50 \mathrm{~cm}$ having 100 turns carries a current of $2.5 \mathrm{~A}$. The magnetic field at one end of the solenoid is
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Verified Answer
The correct answer is:
$3.14 \times 10^{-4} \mathrm{~T}$
Given, length of solenoid, $l=50 \mathrm{~cm}=0.5 \mathrm{~m}$
Number of turns, $N=100$
Current, $I=2.5 \mathrm{~A}$
Number of turns per unit length in the solenoid
$$
n=\frac{N}{l}=\frac{100}{0.5}=200 \text { turns } / \mathrm{m}
$$
Magnetic field at the one end of solenoid,
$$
\begin{aligned}
B & =\frac{\mu_0 n I}{2}=\frac{4 \pi \times 10^{-7} \times 200 \times 2.5}{2} \\
& =10 \pi \times 10^{-5}=\pi \times 10^{-4}=3.14 \times 10^{-4} \mathrm{~T}
\end{aligned}
$$
Number of turns, $N=100$
Current, $I=2.5 \mathrm{~A}$
Number of turns per unit length in the solenoid
$$
n=\frac{N}{l}=\frac{100}{0.5}=200 \text { turns } / \mathrm{m}
$$
Magnetic field at the one end of solenoid,
$$
\begin{aligned}
B & =\frac{\mu_0 n I}{2}=\frac{4 \pi \times 10^{-7} \times 200 \times 2.5}{2} \\
& =10 \pi \times 10^{-5}=\pi \times 10^{-4}=3.14 \times 10^{-4} \mathrm{~T}
\end{aligned}
$$
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