Inductor geometry effect: If the diameter of a coil in an inductor is doubled (other factors like turns and length held constant), how does the inductance change?

Introduction / Context:
Inductance depends on geometry as well as core material. For common air-core solenoids, inductance is roughly proportional to N^2 * A / l, where N is turns, A is cross-sectional area, and l is length. Understanding how size changes alter L is crucial for RF coils and power inductors.

Given Data / Assumptions:

• Coil diameter is doubled.
• Number of turns N and coil length l are unchanged.
• Core properties unchanged.

Concept / Approach:
Area A of the coil cross-section scales with the square of diameter (A ∝ d^2). With L ∝ N^2 * A / l and N, l constant, L ∝ A ∝ d^2. Doubling diameter (d → 2d) makes area four times larger, so inductance increases by a factor of four.

Step-by-Step Solution:
Initial relation: L ∝ N^2 * A / l.A ∝ d^2; N, l fixed.If d → 2d, then A → 4A.Therefore L → 4L (inductance quadruples).

Verification / Alternative check:
Empirical inductor calculators show L scaling roughly with d^2 for a fixed-length, fixed-turn solenoid, aligning with the proportionality argument.

Why Other Options Are Wrong:
(a) only doubles instead of quadruples; (b) and (c) wrongly suggest a decrease in L with larger diameter.

Common Pitfalls:
Forgetting that area grows with diameter squared; confusing changes in diameter with changes in the number of turns.

Final Answer:
Inductance is multiplied by four