Consider a long solenoid of length l and radius r. If n is the number of turns per unit length and µ₀ the permeability of free space, the inductance of the solenoid is:
- A.µ₀π n² r² l✓
- B.µ₀ n² r² l
- C.(µ₀)/(2π)n² r² l
- D.2µ₀π n² r² l
Correct Answer
(A) µ₀π n² r² l
Solution & Explanation
Definition: self-inductance is defined by total flux linkage = L·I, i.e. L=Nφ/I, where φ is the flux through one turn. Step 1 — Field inside a long solenoid. For a long solenoid with n turns per unit length carrying current I, B=µ₀nI. Step 2 — Flux through one turn (cross-section A=πr²). φ=B·A=µ₀nI·πr². Step 3 — Total number of turns over length l. N=n·l. Step 4 — Assemble L=Nφ/I. L=(nl)(µ₀nI·πr²)/I=µ₀n²(πr²)l. The current I cancels, leaving L=µ₀π n² r² l. This matches option A. Note: the structure is L=µ₀n²(Volume)=µ₀n²·(πr²l); the π comes only from the circular cross-section. Options without the π or with extra 2π factors come from mishandling the area or B.
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