The minus sign is an expression of Lenz’s law, indicating that emf opposes the change in current. Units of self-inductance are henries (H) just as for mutual inductance. The larger the self-inductance [latex]L[/latex] of a device, the greater its opposition to any change in current through it. For example, a large coil with many turns and an iron core has a large [latex]L[/latex] and will not allow current to change quickly. To avoid this effect, a small [latex]L[/latex] must be achieved, such as by counterwinding coils as in (Figure).

A 1 H inductor is a large inductor. To illustrate this, consider a device with [latex]L=1\text{.}0 H[/latex] that has a 10 A current flowing through it. What happens if we try to shut off the current rapidly, perhaps in only 1.0 ms? An emf, given by [latex]\text{emf}=-L\left(\Delta I/\Delta t\right)[/latex], will oppose the change. Thus an emf will be induced given by [latex]\text{emf}=-L\left(\Delta I/\Delta t\right)=\left(1\text{.}0 H\right)\left[\left(\text{10 A}\right)/\left(1\text{.}0 ms\right)\right]=\text{10,000 V}[/latex]. The positive sign means this large voltage is in the same direction as the current, opposing its decrease. Such large emfs can cause arcs, damaging switching equipment, and so it may be necessary to change current more slowly.

There are uses for such a large induced voltage. Camera flashes use a battery, two inductors that function as a transformer, and a switching system or oscillator to induce large voltages. (Remember that we need a changing magnetic field, brought about by a changing current, to induce a voltage in another coil.) The oscillator system will do this many times as the battery voltage is boosted to over one thousand volts. (You may hear the high pitched whine from the transformer as the capacitor is being charged.) A capacitor stores the high voltage for later use in powering the flash. (See (Figure).)

It is possible to calculate [latex]L[/latex] for an inductor given its geometry (size and shape) and knowing the magnetic field that it produces. This is difficult in most cases, because of the complexity of the field created. So in this text the inductance [latex]L[/latex] is usually a given quantity. One exception is the solenoid, because it has a very uniform field inside, a nearly zero field outside, and a simple shape. It is instructive to derive an equation for its inductance. We start by noting that the induced emf is given by Faraday’s law of induction as [latex]\text{emf}=-N\left(\Delta \Phi /\Delta t\right)[/latex] and, by the definition of self-inductance, as [latex]\text{emf}=-L\left(\Delta I/\Delta t\right)[/latex]. Equating these yields

Solving for [latex]L[/latex] gives

This equation for the self-inductance [latex]L[/latex] of a device is always valid. It means that self-inductance [latex]L[/latex] depends on how effective the current is in creating flux; the more effective, the greater [latex]\Delta \Phi[/latex]/ [latex]\Delta I[/latex] is.

Let us use this last equation to find an expression for the inductance of a solenoid. Since the area [latex]A[/latex] of a solenoid is fixed, the change in flux is

[latex]\text{Δ}\Phi =\text{Δ}\left(BA\right)=A\text{Δ}B[/latex].

To find
[latex]\text{Δ}B[/latex], we note that the magnetic field of a solenoid is given by [latex]B={\mu }_{0}\text{nI}={\mu }_{0}\frac{\text{NI}}{\ell }[/latex]. (Here [latex]n=N/\ell[/latex], where
[latex]N[/latex] is the number of coils and
[latex]\ell[/latex] is the solenoid’s length.) Only the current changes, so that [latex]\Delta \Phi =A\Delta B={\mu }_{0}\text{NA}\frac{\Delta I}{\ell }[/latex]. Substituting
[latex]\text{Δ}\Phi[/latex] into [latex]L=N\frac{\Delta \Phi }{\Delta I}[/latex] gives

This simplifies to

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

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 (Figure).

where [latex]L[/latex] is the self-inductance of the inductor, and [latex]\Delta I/\Delta t[/latex] is the rate of change of current through it. The minus sign indicates that emf opposes the change in current, as required by Lenz’s law.The unit of self- and mutual inductance is the henry (H), where [latex]1 H=1 \Omega \cdot \text{s}[/latex].The self-inductance [latex]L[/latex] of an inductor is proportional to how much flux changes with current. For an [latex]N[/latex]-turn inductor,

The self-inductance of a solenoid is

where [latex]N[/latex] is its number of turns in the solenoid, [latex]A[/latex] is its cross-sectional area, [latex]\ell[/latex] is its length, and [latex]{\text{μ}}_{0}=4\pi ×{\text{10}}^{\text{−7}}\phantom{\rule{0.25em}{0ex}}\text{T}\cdot \text{m/A}\phantom{\rule{0.10em}{0ex}}[/latex] is the permeability of free space.The energy stored in an inductor [latex]{E}_{\text{ind}}[/latex] is