Question 23.7: Calculate the self-inductance of a 10.0 cm long, 4.00 cm dia...

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

L=\frac{\mu_{0} N^{2} A}{\ell}.                   (23.41)

The cross-sectional area in this example is A=\pi r^{2}=(3.14 \ldots)(0.0200 m )^{2}=1.26 \times 10^{-3} m ^{2}, N is given to be 200, and the length ℓ is 0.100 m. We know the permeability of free space is \mu_{0}=4 \pi \times 10^{-7} T \cdot m / A. Substituting these into the expression for L gives

L=\frac{\left(4 \pi \times 10^{-7} T \cdot m / A \right)(200)^{2}\left(1.26 \times 10^{-3} m ^{2}\right)}{0.100 m }                (23.42)

= 0.632 mH.

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