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A solenoid of $N = 200$ turns, length $L = 20$ cm and cross-section $A = 5\ \text{cm}^2$ has self-inductance approximately:
A126 μH
B1.26 μH
C1.26 mH
D12.6 μH
Answer & Solution
Correct answer: A. 126 μH
$L = \mu_0 N^2 A/L$. Substituting: $L = (4\pi \times 10^{-7})(200)^2 (5\times10^{-4})/(20\times10^{-2}) = (4\pi \times 10^{-7})(40000)(5\times10^{-4})/0.2 = 1.26 \times 10^{-4}$ H $\approx 126\ \mu$H (or equivalently 0.126 mH).
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