Calculating the Self-inductance of a Moderate Size Solenoid

This is a straightforward application of $L=\frac{{\mu }_0{N}^{2}A}{\ell }$, since all quantities in the equation except $L$ are known.

Use the following expression for the self-inductance of a solenoid:

The cross-sectional area in this example is $A={\mathrm{\pi r}}^2=(3\text{.}\text{14}\text{.}\text{.}\text{.})(0\text{.0200 m}{)}^2=1\text{.}\text{26}\times {\text{10}}^{-3}{\text{m}}^2$, $N$ is given to be 200, and the length $\ell$ is 0.100 m. We know the permeability of free space is ${\mu }_0=\mathrm{4\pi }\times {\text{10}}^{\text{−7}}\text{T}\cdot \text{m/A}$. Substituting these into the expression for $L$ gives

This solenoid is moderate in size. Its inductance of nearly a millihenry is also considered moderate.