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A long solenoid has 500 turns. When a current of 2 ampere is passed through it, the resulting magnetic flux linked with each turn of the solenoid is $4 \times 10^{-3} \omega \mathrm{b}$. The selfinductance of the solenoid is
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The correct answer is:
1.0 henry
$$
\begin{aligned}
& N \phi=L i \\
& 500 \times 4 \times 10^{-3}=2 L \\
& L=1.0 \text { henry }
\end{aligned}
$$
\begin{aligned}
& N \phi=L i \\
& 500 \times 4 \times 10^{-3}=2 L \\
& L=1.0 \text { henry }
\end{aligned}
$$
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