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A long solenoid has 500 turns, when a current of $2 \mathrm{~A}$ is passed through it, the resulting magnetic flux linked with each turn of the solenoid is $4 \times 10^{-3} \mathrm{~Wb}$, then selfinduction of the solenoid is
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Verified Answer
The correct answer is:
$1.0 \mathrm{H}$
Given, number of turns in the solenoid,
$$
N=500
$$
Current, $I=2 \mathrm{~A}$
Magnetic flux, $\phi=4 \times 10^{-3} \mathrm{~Wb}$
Self-inductance of coil,
$$
L=\frac{N \phi}{I}=\frac{500 \times 4 \times 10^{-3}}{2}=1 \mathrm{H}
$$
$$
N=500
$$
Current, $I=2 \mathrm{~A}$
Magnetic flux, $\phi=4 \times 10^{-3} \mathrm{~Wb}$
Self-inductance of coil,
$$
L=\frac{N \phi}{I}=\frac{500 \times 4 \times 10^{-3}}{2}=1 \mathrm{H}
$$
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