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A solenoid of 1200 turns is wound uniformly in a single layer on a glass tube $2 \mathrm{~m}$ long and $0.2 \mathrm{~m}$ in diameter. The magnetic intensity at the center of the solenoid when a current of 2 A flows through it is:
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Verified Answer
The correct answer is:
$1.2 \times 10^3 \mathrm{Am}^{-1}$
Given that number of turns, $n=1200$
Current, $\mathrm{I}=2 \mathrm{~A}$
Magnetic field at centre inside the solenoid is given by,
$\mathrm{B}=\mu_0 \mathrm{nI}$
So magnetic intensity at centre of the solenoid,
$\begin{aligned}
& \mathrm{H}=\frac{\mathrm{B}}{\mu_0}=\mathrm{nI}=\left(\frac{1200}{2}\right)(2) \quad\left(\because \mathrm{B}=\mu_0 \mathrm{H}\right) \\
& \mathrm{H}=1.2 \times 10^3 \mathrm{Am}^{-1}
\end{aligned}$
Current, $\mathrm{I}=2 \mathrm{~A}$
Magnetic field at centre inside the solenoid is given by,
$\mathrm{B}=\mu_0 \mathrm{nI}$
So magnetic intensity at centre of the solenoid,
$\begin{aligned}
& \mathrm{H}=\frac{\mathrm{B}}{\mu_0}=\mathrm{nI}=\left(\frac{1200}{2}\right)(2) \quad\left(\because \mathrm{B}=\mu_0 \mathrm{H}\right) \\
& \mathrm{H}=1.2 \times 10^3 \mathrm{Am}^{-1}
\end{aligned}$
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