Self-inductance versus turns – doubling the turns of a coil A coil with 200 turns has a self-inductance of 10 mH. If the number of turns is doubled to 400 while all other factors remain unchanged (same core and geometry), what will be the new self-inductance?

Introduction:
Inductance quantifies a coil’s ability to store magnetic energy for a given current. For a given magnetic circuit (same core material and geometry), the self-inductance depends strongly on the number of turns. This relationship is routinely used when scaling inductors and transformer windings.

Given Data / Assumptions:

• Initial turns N1 = 200 with L1 = 10 mH.
• Final turns N2 = 400 (doubled).
• Core permeability, cross-section, mean length, and winding placement unchanged.

Concept / Approach:

For a fixed magnetic circuit, inductance varies with the square of the number of turns: L ∝ N^2. Therefore, doubling the turns increases L by a factor of 4. This assumes the core is not saturating and leakage or proximity effects remain negligible under the small-signal assumption used to define inductance.

Step-by-Step Solution:

Verification / Alternative check:

Energy check: stored energy W = 0.5 * L * I^2. With unchanged current and quadrupled L, energy scales accordingly, consistent with stronger magnetizing MMF (NI) for the same current.

Why Other Options Are Wrong:

2.5 mH and 5 mH suggest inverse or linear scaling; 20 mH suggests only doubling; 80 mH suggests octupling. The correct square-law gives 40 mH.

Common Pitfalls:

Forgetting the N^2 dependence; assuming linear dependence on N; ignoring real-world parasitics (important at high frequency but not for this ideal calculation).

Final Answer:

40 mH