Electrostatics

Consider a circular coil of N turns and radius R.
Let a current I be flowing in the coil.
Magnetic field, $B = \frac{\mu_{0}NI}{2R}$
Flux linkage, $N\phi = N[BA]$
$N\phi = N \left[\frac{\mu_{0}NI(\pi R^{2})}{2R}\right]$
$N\phi = \frac{\mu_{0}N^{2}I\pi R}{2}$
Self Inductance $L = \frac{N\phi}{I} = \frac{\mu_{0}N^{2}\pi R}{2}$
Step Solution:
Step 1: Identify the magnetic field produced at the center of the circular coil using $B = \frac{\mu_{0}NI}{2R}$.
Step 2: Calculate the total magnetic flux linkage $N\phi$ by multiplying the number of turns N, the magnetic field B, and the cross-sectional area $A = \pi R^{2}$.
Step 3: Substitute $N\phi$ into the self-inductance formula $L = \frac{N\phi}{I}$ to cancel out the current I and obtain the final expression $L = \frac{\mu_{0}N^{2}\pi R}{2}$.