Question

A circular coil of radius $R$ has $N$ turns of a wire. The coefficient of self-induction of the coil will be ($\mu_0 = $ permeability of free space)

• A. $\frac{\mu_0 N \pi R^2}{2}$
• B. $\frac{\mu_0 N \pi R}{4}$
• C. $\frac{\mu_0 N^2 \pi R}{2}$
• D. $\frac{\mu_0 N \pi R^2}{4}$

Hint:

Always remember a fundamental structural rule: the self-inductance ($L$) of any inductor coil configuration is always directly proportional to the square of the number of turns ($L \propto N^2$). Looking at the choices, only option (C) contains an $N^2$ term, allowing you to instantly select the correct formula without executing any derivations!

Answer 1

The Correct Option is C

Step 1: Understanding the Question:
Find the self-inductance L of a flat circular coil with N turns and radius R.

Step 2: Key Formula or Approach:
Magnetic field at the coil center: B = μ₀NI/(2R). Total flux linkage: Φ = N·B·A with A = πR². Self-inductance is defined by Φ = L·I, so L = Φ/I.

Step 3: Detailed Explanation:
Substitute B and A: Φ = N·(μ₀NI/(2R))·(πR²) = (μ₀N²πR²I)/(2R) = (μ₀N²πRI)/2. Dividing by I yields L = (μ₀N²πR)/2.

Step 4: Final Answer:
The self-inductance is (μ₀N²πR)/2, option (C).