Inductance scaling with number of turns for a fixed-geometry single-layer air-core coil A single-layer air-core coil has n turns and inductance L. If a new coil is wound with 2n turns while keeping length and diameter the same, what is the new inductance?

Introduction / Context:
The inductance of a coil depends on geometry and the square of the number of turns. For small changes that do not alter length or diameter, the proportionality L ∝ n^2 is a reliable design rule, especially for air-core coils at low frequencies where parasitics are minimal.

Given Data / Assumptions:

• Single-layer, air-core coil.
• Coil length and diameter held constant.
• Parasitic capacitance and proximity effects neglected.

Concept / Approach:
For fixed geometry (area A and magnetic path length l roughly constant), L ∝ n^2. If the number of turns is doubled from n to 2n, the inductance becomes L_new = (2n)^2 / n^2 * L = 4L.

Step-by-Step Solution:
Start with L ∝ n^2 for fixed geometry.Replace n → 2n.Compute scaling: (2n)^2 = 4 n^2.Hence L_new = 4 L.

Verification / Alternative check:
Textbook solenoid formula L = μ0 n^2 A / l directly shows the n^2 dependence when A and l are unchanged.

Why Other Options Are Wrong:
0.5L, L, and 2L contradict the n^2 law; doubling turns cannot yield linear scaling in L for a fixed geometry.

Common Pitfalls:
Forgetting that changes in coil dimensions can also affect L; at very high frequency, proximity and skin effects alter effective parameters, but the basic low-frequency relation remains n^2.

Final Answer:
4 L