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Self-inductance of a long solenoid of length l, cross-section A, with n turns per unit length is
AL = μ₀ A / (n² l)
BL = μ₀ n² A l
CL = μ₀ n A / l
DL = μ₀ n l / A
Answer & Solution
Correct answer: B. L = μ₀ n² A l
B = μ₀nI inside; Φ_B per turn = μ₀nI·A; total linkage N Φ = (nl)(μ₀nIA) = μ₀n²AlI. So L = μ₀n²Al.
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