30.6.1 Calculating the Induced Field

The induced electric field is peculiar in another way: It is nonconservative. Recall that a force is conservative if it does no net work on a particle moving around a closed path. “Uphills” are balanced by “downhills.” We can associate a potential energy with a conservative force, hence we have gravitational potential energy for the conservative gravitational force and electric potential energy for the conservative electric force of charges (a Coulomb electric field).

But a charge moving around a closed path in the induced electric field of Figure 30.30 is always being pushed in the same direction by the electric force $\vec{F} = q \vec{E}$ . There’s never any negative work to balance the positive work, so the net work done in going around a closed path is not zero. Because it’s nonconservative, we cannot associate an electric potential with an induced electric field. Only the Coulomb field of charges has an electric potential.

However, we can associate the induced field with the emf of Faraday’s law. The emf was defined as the work required per unit charge to separate the charge. That is,

ε = \frac{W}{q}

(30.20)

In batteries, a familiar source of emf, this work is done by chemical forces. But the emf that appears in Faraday’s law arises when work is done by the force of an induced electric field.

A photograph of a mobile phone over a wireless charging pad. — It’s likely that you’ve charged your phone—or seen others doing so—simply by placing it on a flat charging pad. This is *inductive charging*, which makes use of Faraday’s law. The base unit drives an oscillating current through a large, flat coil, generating an oscillating magnetic field perpendicular to the flat pad. The changing flux through a smaller coil inside your phone then induces a current that recharges the battery. The widely used Qi standard (the Chinese word for “vital energy,” pronounced *chee*) uses a frequency of 110–250 KHz for a 5 W system and a higher frequency of 300 kHz for newer 15 W fast-charging systems. The efficiency—the ratio of charging power to consumed power—is only about 50%, considerably less than the near-perfect efficiency of a plug-in charger.

If a charge q moves through a small displacement $d \vec{s}$ , the small amount of work done by the electric field is $d W = \vec{F} \cdot d \vec{s} = q \vec{E} \cdot d \vec{s}$ . The emf of Faraday’s law is an emf around a closed curve through which the magnetic flux $Φ_{m}$ is changing. The work done by the induced electric field as charge q moves around a closed curve is

W_{closed curve} = q \oint \vec{E} \cdot d \vec{s}

(30.21)

where the integration symbol with the circle is the same as the one we used in Ampère’s law to indicate an integral around a closed curve. If we use this work in Equation 30.20, we find that the emf around a closed loop is

ε = \frac{W_{closed curve}}{q} = \oint \vec{E} \cdot d \vec{s}

(30.22)

If we restrict ourselves to situations such as Figure 30.30 where the loop is perpendicular to the magnetic field and only the field is changing, we can write Faraday’s law as $ε = | d Φ_{m} / d t | = A | d B / d t |$ . Consequently

\oint \vec{E} \cdot d \vec{s} = A | \frac{d B}{d t} |

(30.23)

Equation 30.23 is an alternative statement of Faraday’s law that relates the induced electric field to the changing magnetic field.

The solenoid in FIGURE 30.32a provides a good example of the connection between $\vec{E}$ and $\vec{B}$ . If there were a conducting loop inside the solenoid, we could use Lenz’s law to determine that the direction of the induced current would be clockwise. But Faraday’s law, in the form of Equation 30.23, tells us that an induced electric field is present whether there’s a conducting loop or not. The electric field is induced simply due to the fact that $\vec{B}$ is changing.

The figure labeled (“a”) illustrates an increasing magnetic field inside a solenoid.

The figure labeled (b) illustrates the induced electric field around the magnetic field lines.

Full Alternative Text

The shape and direction of the induced electric field have to be such that it could drive a current around a conducting loop, if one were present, and it has to be consistent with the cylindrical symmetry of the solenoid. The only possible choice, shown in FIGURE 30.32b, is an electric field that circulates clockwise around the magnetic field lines.

NOTE Circular electric field lines violate the Chapter 23 rule that electric field lines have to start and stop on charges. However, that rule applied only to Coulomb fields created by source charges. An induced electric field is a non-Coulomb field created not by source charges but by a changing magnetic field. Without source charges, induced electric field lines must form closed loops.

To use Faraday’s law, choose a clockwise circle of radius r as the closed curve for evaluating the integral. FIGURE 30.32c shows that the electric field vectors are everywhere tangent to the curve, so the line integral of $\vec{E}$ is

\oint \vec{E} \cdot d \vec{s} = E l = 2 π r E

(30.24)

where $l = 2 π r$ is the length of the closed curve. This is exactly like the integrals we did for Ampère’s law in Chapter 29.

If we stay inside the solenoid $(r < R)$ , the flux passes through area $A = π r^{2}$ and Equation 30.24 becomes

\oint \vec{E} \cdot d \vec{s} = 2 π r E = A | \frac{d B}{d t} | = π r^{2} | \frac{d B}{d t} |

(30.25)

Thus the strength of the induced electric field inside the solenoid is

E_{inside} = \frac{r}{2} | \frac{d B}{d t} |

(30.26)

This result shows very directly that the induced electric field is created by a changing magnetic field. A constant $\vec{B}$ , with $d B / d t = 0$ , would give $E = 0$ .

EXAMPLE 30.10 An induced electric field

A 4.0-cm-diameter solenoid is wound with 2000 turns per meter. The current through the solenoid oscillates at 60 Hz with an amplitude of 2.0 A. What is the maximum strength of the induced electric field inside the solenoid?

MODEL Assume that the magnetic field inside the solenoid is uniform.

VISUALIZE The electric field lines are concentric circles around the magnetic field lines, as was shown in Figure 30.32b. They reverse direction twice every period as the current oscillates.

SOLVE You learned in Chapter 29 that the magnetic field strength inside a solenoid with n turns per meter is $B = μ_{0} n I$ . In this case, the current through the solenoid is $I = I_{0} sin ω t$ , where $I_{0} = 2.0 A$ is the peak current and $ω = 2 π (60 Hz) = 377 rad/s$ . Thus the induced electric field strength at radius r is

E = \frac{r}{2} | \frac{d B}{d t} | = \frac{r}{2} \frac{d}{d t} (μ_{0} n I_{0} sin ω t) = \frac{1}{2} μ_{0} n r ω I_{0} cos ω t

The field strength is maximum at maximum radius $(r = R)$ and at the instant when $cos ω t = 1$ . That is,

E_{max} = \frac{1}{2} μ_{0} n R ω I_{0} = 0.019 V / m

REVIEW This field strength, although not large, is similar to the field strength that the emf of a battery creates in a wire. Hence this induced electric field can drive a substantial induced current through a conducting loop if a loop is present. But the induced electric field exists inside the solenoid whether or not there is a conducting loop.

Occasionally it is useful to have a version of Faraday’s law without the absolute value signs. The essence of Lenz’s law is that the emf $ε$ opposes the change in $Φ_{m}$ . Mathematically, this means that $ε$ must be opposite in sign to $d B / d t$ . Consequently, we can write Faraday’s law as

ε = \oint \vec{E} \cdot d \vec{s} = - \frac{d Φ_{m}}{d t}

(30.27)

For practical applications, it’s always easier to calculate just the magnitude of the emf with Faraday’s law and to use Lenz’s law to find the direction of the emf or the induced current. However, the mathematically rigorous version of Faraday’s law in Equation 30.27 will prove to be useful when we combine it with other equations, in Chapter 31, to predict the existence of electromagnetic waves.