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The self inductance of a coil having 400 turns is $10 \mathrm{mH}$. The magnetic flux through the cross section of the coil corresponding to current $2 \mathrm{~mA}$ is
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Verified Answer
The correct answer is:
$2 \times 10^{-5} \mathrm{~Wb}$
Here, $N=400, L=10 \mathrm{mH}=10 \times 10^{-3} \mathrm{H}$ $I=2 \mathrm{~mA}=2 \times 10^{-3} \mathrm{~A}$
Total magnetic flux linked with the coil,
$$
\begin{aligned}
\phi=N L I & =400 \times\left(10 \times 10^{-3}\right) \times 2 \times 10^{-3} \\
& =8 \times 10^{-3} \mathrm{~Wb}
\end{aligned}
$$
Magnetic flux through the cross-section of the coil = Magnetic flux linked with each turn
$$
=\frac{\phi}{N}=\frac{8 \times 10^{-3}}{400}=2 \times 10^{-5} \mathrm{~Wb}
$$
Total magnetic flux linked with the coil,
$$
\begin{aligned}
\phi=N L I & =400 \times\left(10 \times 10^{-3}\right) \times 2 \times 10^{-3} \\
& =8 \times 10^{-3} \mathrm{~Wb}
\end{aligned}
$$
Magnetic flux through the cross-section of the coil = Magnetic flux linked with each turn
$$
=\frac{\phi}{N}=\frac{8 \times 10^{-3}}{400}=2 \times 10^{-5} \mathrm{~Wb}
$$
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