# 23.9 Inductance  (Page 3/10)

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$L=N\frac{\Delta \Phi }{\Delta I}=N\frac{{\mu }_{0}\text{NA}\frac{\Delta I}{\ell }}{\Delta I}\text{.}$

This simplifies to

$L=\frac{{\mu }_{0}{N}^{2}A}{\ell }\text{(solenoid).}$

This is the self-inductance of a solenoid of cross-sectional area $A$ and length $\ell$ . Note that the inductance depends only on the physical characteristics of the solenoid, consistent with its definition.

## 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:

$L=\frac{{\mu }_{0}{N}^{2}A}{\ell }\text{.}$

The cross-sectional area in this example is $A={\mathrm{\pi r}}^{2}=\left(3\text{.}\text{14}\text{.}\text{.}\text{.}\right)\left(0\text{.0200 m}{\right)}^{2}=1\text{.}\text{26}×{\text{10}}^{-3}\phantom{\rule{0.25em}{0ex}}{\text{m}}^{2}$ , $N$ is given to be 200, and the length $\ell$ is 0.100 m. We know the permeability of free space is ${\mu }_{0}=4\pi ×{\text{10}}^{\text{−7}}\phantom{\rule{0.25em}{0ex}}\text{T}\cdot \text{m/A}$ . Substituting these into the expression for $L$ gives

$\begin{array}{lll}L& =& \frac{\left(4\pi ×{\text{10}}^{-7}\phantom{\rule{0.25em}{0ex}}\text{T}\cdot \text{m/A}\right)\left(\text{200}{\right)}^{2}\left(1.26×{\text{10}}^{-3}\phantom{\rule{0.25em}{0ex}}{\text{m}}^{2}\right)}{0.100 m}\\ & =& 0\text{.}\text{632 mH}\text{.}\end{array}$

Discussion

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

One common application of inductance is used in traffic lights that can tell when vehicles are waiting at the intersection. An electrical circuit with an inductor is placed in the road under the place a waiting car will stop over. The body of the car increases the inductance and the circuit changes sending a signal to the traffic lights to change colors. Similarly, metal detectors used for airport security employ the same technique. A coil or inductor in the metal detector frame acts as both a transmitter and a receiver. The pulsed signal in the transmitter coil induces a signal in the receiver. The self-inductance of the circuit is affected by any metal object in the path. Such detectors can be adjusted for sensitivity and also can indicate the approximate location of metal found on a person. (But they will not be able to detect any plastic explosive such as that found on the “underwear bomber.”) See [link] .

## Energy stored in an inductor

We know from Lenz’s law that inductances oppose changes in current. There is an alternative way to look at this opposition that is based on energy. Energy is stored in a magnetic field. It takes time to build up energy, and it also takes time to deplete energy; hence, there is an opposition to rapid change. In an inductor, the magnetic field is directly proportional to current and to the inductance of the device. It can be shown that the energy stored in an inductor     ${E}_{\text{ind}}$ is given by

${E}_{\text{ind}}=\frac{1}{2}{\text{LI}}^{2}\text{.}$

This expression is similar to that for the energy stored in a capacitor.

## 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

$\begin{array}{lll}{E}_{\text{ind}}& =& \frac{1}{2}{\text{LI}}^{2}\\ & =& 0.5\left(0.632×{\text{10}}^{-3}\phantom{\rule{0.25em}{0ex}}\text{H}\right)\left(\text{30.0 A}{\right)}^{2}=\text{0.284 J}\text{.}\end{array}$

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.

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