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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 $4 \times 10^{-3} \mathrm{~Wb}$. The self-inductance of the solenoid is
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The correct answer is:
$1.0 \mathrm{H}$
Key Idea : Inductance of a coil is numerically equal to the emf induced in the coil when the current in the coil changes at the rate of $1 \mathrm{As}^{-1}$. If $I$ is the current flowing in the circuit, then flux linked with the circuit is observed to be proportional to $I$, ie,
$\begin{aligned}
& \phi \propto I \\
& \text{or } \phi=L I
\end{aligned}$
where $L$ is called the self-inductance or coefficient of self-inductance or simply inductance of the coil.
Net flux through solenoid,
$\phi=500 \times 4 \times 10^{-3}=2 \mathrm{~Wb}$
or $2=L \times 2$ [after putting values in Eq. or $L=1 \mathrm{H}$
$\begin{aligned}
& \phi \propto I \\
& \text{or } \phi=L I
\end{aligned}$
where $L$ is called the self-inductance or coefficient of self-inductance or simply inductance of the coil.
Net flux through solenoid,
$\phi=500 \times 4 \times 10^{-3}=2 \mathrm{~Wb}$
or $2=L \times 2$ [after putting values in Eq. or $L=1 \mathrm{H}$
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