14.2 Self-Inductance and Inductors

Inductors

Figure 14.5 shows some of the magnetic field lines due to the current in a circular loop of wire. If the current is constant, the magnetic flux through the loop is also constant. However, if the current I were to vary with time—say, immediately after switch S is closed—then the magnetic flux ${\text{Φ}}_{\text{m}}$ would correspondingly change. Then Faraday’s law tells us that an emf $\epsilon$ would be induced in the circuit, where

$$\epsilon =-\frac{d{\text{Φ}}_{\text{m}}}{dt}.$$

Since the magnetic field due to a current-carrying wire is directly proportional to the current, the flux due to this field is also proportional to the current; that is,

$${\text{Φ}}_{\text{m}}\propto I.$$

This can also be written as

$${\text{Φ}}_{\text{m}}=LI$$

where the constant of proportionality L is known as the self-inductance of the wire loop. If the loop has N turns, this equation becomes

By convention, the positive sense of the normal to the loop is related to the current by the right-hand rule, so in Figure 14.5, the normal points downward. With this convention, ${\text{Φ}}_{\text{m}}$ is positive in Equation 14.9, so L always has a positive value.

For a loop with N turns, $\epsilon =\text{−}Nd{\text{Φ}}_{\text{m}}\text{/}dt,$ so the induced emf may be written in terms of the self-inductance as

When using this equation to determine L, it is easiest to ignore the signs of $\epsilon \text{and}dI\text{/}dt,$ and calculate L as

$$L=\frac{\left|\epsilon \right|}{\left|dI\text{/}dt\right|}.$$

Since self-inductance is associated with the magnetic field produced by a current, any configuration of conductors possesses self-inductance. For example, besides the wire loop, a long, straight wire has self-inductance, as does a coaxial cable. A coaxial cable is most commonly used by the cable television industry and may also be found connecting to your cable modem. Coaxial cables are used due to their ability to transmit electrical signals with minimal distortions. Coaxial cables have two long cylindrical conductors that possess current and a self-inductance that may have undesirable effects.

A circuit element used to provide self-inductance is known as an inductor. It is represented by the symbol shown in Figure 14.6, which resembles a coil of wire, the basic form of the inductor. Figure 14.7 shows several types of inductors commonly used in circuits.

In accordance with Lenz’s law, the negative sign in Equation 14.10 indicates that the induced emf across an inductor always has a polarity that opposes the change in the current. For example, if the current flowing from A to B in Figure 14.8(a) were increasing, the induced emf (represented by the imaginary battery) would have the polarity shown in order to oppose the increase. If the current from A to B were decreasing, then the induced emf would have the opposite polarity, again to oppose the change in current (Figure 14.8(b)). Finally, if the current through the inductor were constant, no emf would be induced in the coil.

One common application of inductance is to allow traffic signals to sense when vehicles are waiting at a street intersection. An electrical circuit with an inductor is placed in the road underneath the location where a waiting car will stop. 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 from the transmitter coil induces a signal in the receiver. The self-inductance of the circuit is affected by any metal object in the path (Figure 14.9). Metal detectors can be adjusted for sensitivity and can also sense the presence of metal on a person.

Large induced voltages are found in camera flashes. Camera flashes use a battery, two inductors that function as a transformer, and a switching system or oscillator to induce large voltages. Recall from Oscillations on oscillations that “oscillation” is defined as the fluctuation of a quantity, or repeated regular fluctuations of a quantity, between two extreme values around an average value. Also recall (from Electromagnetic Induction on electromagnetic induction) that we need a changing magnetic field, brought about by a changing current, to induce a voltage in another coil. The oscillator system does this many times as the battery voltage is boosted to over 1000 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.

A good approach for calculating the self-inductance of an inductor consists of the following steps:

To demonstrate this procedure, we now calculate the self-inductances of two inductors.

Cylindrical Solenoid

Consider a long, cylindrical solenoid with length l, cross-sectional area A, and N turns of wire. We assume that the length of the solenoid is so much larger than its diameter that we can take the magnetic field to be $B={\mu }_{0}nI$ throughout the interior of the solenoid, that is, we ignore end effects in the solenoid. With a current I flowing through the coils, the magnetic field produced within the solenoid is

$$B={\mu }_0\left(\frac{N}{l}\right)I,$$

so the magnetic flux through one turn is

$${\text{Φ}}_{\text{m}}=BA=\frac{{\mu }_{0}NA}{l}I.$$

Using Equation 14.9, we find for the self-inductance of the solenoid,

If $n=N\text{/}l$ is the number of turns per unit length of the solenoid, we may write Equation 14.13 as

where $V=Al$ is the volume of the solenoid. Notice that the self-inductance of a long solenoid depends only on its physical properties (such as the number of turns of wire per unit length and the volume), and not on the magnetic field or the current. This is true for inductors in general.

Rectangular Toroid

A toroid with a rectangular cross-section is shown in Figure 14.10. The inner and outer radii of the toroid are $R_1\text{and}{R}_2,\text{and}h$ is the height of the toroid. Applying Ampère’s law in the same manner as we did in Example 2 in Section 13.4 for a toroid with a circular cross-section, we find the magnetic field inside a rectangular toroid is also given by

$$B=\frac{{\mu }_{0}NI}{2\pi r},$$

where r is the distance from the central axis of the toroid. Because the field changes within the toroid, we must calculate the flux by integrating over the toroid’s cross-section. Using the infinitesimal cross-sectional area element $da=hdr$ shown in Figure 14.10, we obtain

Now from Equation 14.16, we obtain for the self-inductance of a rectangular toroid

$$L=\frac{N{\text{Φ}}_{\text{m}}}{I}=\frac{{\mu }_0{N}^{2}h}{2\pi }\text{ln}\frac{R_2}{R_1}.$$

As expected, the self-inductance is a constant determined by only the physical properties of the toroid.

Summary

• Current changes in a device induce an emf in the device itself, called self-inductance,

  $$\epsilon =\text{−}L\frac{dI}{dt},$$

  where L is the self-inductance of the inductor and $dI\text{/}dt$ 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-inductance and inductance is the henry (H), where $1\text{H}=1\text{Ω}·\text{s}$.
• The self-inductance of a solenoid is

  $$L=\frac{{\mu }_0{N}^{2}A}{l},$$

  where N is its number of turns in the solenoid, A is its cross-sectional area, l is its length, and ${\mu }_0=4\pi \times {10}^{\mathrm{-7}}\text{T}·\text{m/A}$ is the permeability of free space.
• The self-inductance of a toroid is

  $$L=\frac{{\mu }_0{N}^{2}h}{2\pi }\text{ln}\frac{R_2}{R_1},$$

  where N is its number of turns in the toroid, $R_1\text{and}{R}_2$ are the inner and outer radii of the toroid, h is the height of the toroid, and ${\mu }_0=4\pi \times {10}^{\mathrm{-7}}\text{T}·\text{m/A}$ is the permeability of free space.

Conceptual Questions

Problems

The current shown in part (a) below is increasing, whereas that shown in part (b) is decreasing. In each case, determine which end of the inductor is at the higher potential.

The current I(t) through a 5.0-mH inductor varies with time, as shown below. The resistance of the inductor is $5.0\text{Ω}.$ Calculate the voltage across the inductor at $$t=2.0\text{ms},t=4.0\text{ms},\text{and}t=8.0\text{ms}$$.

Suppose that a rectangular toroid has 2000 windings and a self-inductance of 0.040 H. If $h=0.10\text{m}$, what is the ratio of its outer radius to its inner radius?