11.9 Inductance  (Page 3/10)

Calculating the self-inductance of a moderate size solenoid

Calculate the self-inductance of a 10.0 cm long, 4.00 cm diameter solenoid that has 200 coils.

Strategy

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

Solution

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

Discussion

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

Calculating the energy stored in the field of a solenoid

How much energy is stored in the 0.632 mH inductor of the preceding example when a 30.0 A current flows through it?

Strategy

The energy is given by the equation $E_{\text{ind}}=\frac{1}{2}{\text{LI}}^2$ , and all quantities except $E_{\text{ind}}$ are known.

Solution

Substituting the value for $L$ found in the previous example and the given current into $E_{\text{ind}}=\frac{1}{2}{\text{LI}}^2$ gives

Discussion

This amount of energy is certainly enough to cause a spark if the current is suddenly switched off. It cannot be built up instantaneously unless the power input is infinite.