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Q.
The self inductance of a solenoid that has a cross-sectional area of $1\, cm ^{2}$, a length of 0.10 m and 1000 turns of wire is
Electromagnetic Induction
Solution:
$L =\frac{\mu_{0} N^{2} A}{l}=\frac{\left(4 \pi \times 10^{-7} \frac{ H }{ m }\right)(1000)^{2}\left(10^{-4} m ^{2}\right)}{(0.10 m )}$
$=1.26 \times 10^{-3} H =1.26\, mH$