30.8.1 The Potential Difference Across an Inductor

An inductor is not very interesting when the current through it is steady. If the inductor is ideal, with R=0 Ω, the potential difference due to a steady current is zero. Inductors become important circuit elements when currents are changing. FIGURE 30.38a shows a steady current into the left side of an inductor. The solenoid’s magnetic field passes through the coils of the solenoid, establishing a flux.

FIGURE 30.38 Increasing the current through an inductor.

The figure labeled (“a”) illustrates a steady current into an inductor.
The figure labeled (b) illustrates an increasing current into an inductor.
Full Alternative Text

In FIGURE 30.38b, the current into the solenoid is increasing. This creates an increasing flux to the left. According to Lenz’s law, an induced current in the coils will oppose this increase by creating an induced magnetic field pointing to the right. This requires the induced current to be opposite the current into the solenoid. This induced current will carry positive charge carriers to the left until a potential difference is established across the solenoid.

You saw a similar situation in Section 30.2. The induced current in a conductor moving through a magnetic field carried positive charge carriers to the top of the wire and established a potential difference across the conductor. The induced current in the moving wire was due to magnetic forces on the moving charges. Now, in Figure 30.38b, the induced current is due to the non-Coulomb electric field induced by the changing magnetic field. Nonetheless, the outcome is the same: a potential difference across the conductor.

We can use Faraday’s law to find the potential difference. The emf induced in a coil is

εcoil=N|dΦper turndt|=|dΦmdt| (30.34)

where Φm=NΦper turn is the total flux through all the coils. The inductance was defined such that Φm=LI, so Equation 30.34 becomes

εcoil=L|dIdt| (30.35)

The induced emf is directly proportional to the rate of change of current through the coil. We’ll consider the appropriate sign in a moment, but Equation 30.35 gives us the size of the potential difference that is developed across a coil as the current through the coil changes. Note that εcoil=0 for a steady, unchanging current.

FIGURE 30.39 shows the same inductor, but now the current (still in to the left side) is decreasing. To oppose the decrease in flux, the induced current is in the same direction as the input current. The induced current carries charge to the right and establishes a potential difference opposite that in Figure 30.38b.

FIGURE 30.39 Decreasing the current through an inductor.

The figure illustrates a decreasing current into an inductor.
FIGURE 30.39 Full Alternative Text

NOTE   Notice that the induced current does not oppose the current through the inductor, which is from left to right in both Figures 30.38 and 30.39. Instead, in accordance with Lenz’s law, the induced current opposes the change in the current in the solenoid. The practical result is that it is hard to change the current through an inductor. Any effort to increase or decrease the current is met with opposition in the form of an opposing induced current. You can think of the current in an inductor as having inertia, trying to continue what it was doing without change.

Before we can use inductors in a circuit we need to establish a rule about signs that is consistent with our earlier circuit analysis. FIGURE 30.40 first shows current I passing through a resistor. You learned in Chapter 28 that the potential difference across a resistor is ΔVres=−ΔVR=−IR, where the minus sign indicates that the potential decreases in the direction of the current.

FIGURE 30.40 The potential difference across a resistor and an inductor.

The figure illustrates the potential differences across a resistor and an inductor.
FIGURE 30.40 Full Alternative Text

We’ll use the same convention for an inductor. The potential difference across an inductor, measured along the direction of the current, is

ΔVL=−LdIdt (30.36)

If the current is increasing (dI/dt>0), the input side of the inductor is more positive than the output side and the potential decreases in the direction of the current (ΔVL<0). This was the situation in Figure 30.38b. If the current is decreasing (dI/dt<0), the input side is more negative and the potential increases in the direction of the current (ΔVL>0). This was the situation in Figure 30.39.

The potential difference across an inductor can be very large if the current changes very abruptly (large dI/dt). FIGURE 30.41 shows an inductor connected across a battery. There is a large current through the inductor, limited only by the internal resistance of the battery. Suppose the switch is suddenly opened. A very large induced voltage is created across the inductor as the current rapidly drops to zero. This potential difference (plus ΔVbat ) appears across the gap of the switch as it is opened. A large potential difference across a small gap often creates a spark.

FIGURE 30.41 Creating sparks.

The figure consists of two parts that illustrate the circuit diagrams of a spark at the switch connected to an inductor.
FIGURE 30.41 Full Alternative Text

Indeed, this is exactly how the spark plugs in your car work. The car’s generator sends a current through the coil, which is a big inductor. When a switch is suddenly opened, breaking the current, the induced voltage, typically a few thousand volts, appears across the terminals of the spark plug, creating the spark that ignites the gasoline. Older cars use a distributor to open and close an actual switch; more recent cars have electronic ignition in which the mechanical switch has been replaced by a transistor.

EXAMPLE 30.13 Large voltage across an inductor

A 1.0 A current passes through a 10 mH inductor coil. What potential difference is induced across the coil if the current drops to zero in 5.0 μs?

MODEL Assume this is an ideal inductor, with R=0 Ω, and that the current decrease is linear with time.

SOLVE The rate of current decrease is

dIdt≈ΔIΔt=−1.0 A5.0×10−6 s=−2.0×105 A/s

The induced voltage is

ΔVL=−LdIdt≈−(0.010 H)(−2.0×105 A/s)=2000 V

REVIEW Inductors may be physically small, but they can pack a punch if you try to change the current through them too quickly.