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:
The problem requires finding the analytical expression for the coefficient of self-induction ($L$) of a flat, closely-wound circular coil containing a total of $N$ turns and a cross-sectional radius $R$.

Step 2: Key Formula or Approach:
1. The magnetic field ($B$) at the geometric center of a circular coil carrying a current $I$ is given by: $$B = \frac{\mu_0 N I}{2R}$$ 2. The total magnetic flux linkage ($\Phi$) passing through all $N$ loops of the coil area $A = \pi R^2$ is: $$\Phi = N \cdot (B \cdot A)$$ 3. The self-inductance ($L$) is defined by the fundamental relationship matching total flux per unit current: $$\Phi = L \cdot I \implies L = \frac{\Phi}{I}$$

Step 3: Detailed Explanation:
Let's first compute the total magnetic flux linked across the entire multi-turn coil assembly. Substitute the magnetic field expression $B$ and area $A = \pi R^2$ into the flux linkage equation: $$\Phi = N \cdot \left( \frac{\mu_0 N I}{2R} \right) \cdot \left( \pi R^2 \right)$$ Combine the terms together: $$\Phi = \frac{\mu_0 N^2 \pi R^2 I}{2R}$$ One factor of the radius $R$ cancels out between the numerator and denominator: $$\Phi = \frac{\mu_0 N^2 \pi R I}{2}$$ Now, to find the self-inductance coefficient $L$, divide the total flux linkage expression by the current variable $I$: $$L = \frac{\Phi}{I} = \frac{\mu_0 N^2 \pi R}{2}$$

Step 4: Final Answer:
The coefficient of self-induction equals $\frac{\mu_0 N^2 \pi R}{2}$, which matches option (C).