Ordinary Coulomb electric fields are created by charges, but a second way to create an electric field is by having a changing magnetic field. That’s Faraday’s law. Ordinary magnetic fields are created by currents, but now we see that a second way to create a magnetic field is by having a changing electric field. Just as the electric field created by a changing $\vec{B}$ is called an induced electric field, the magnetic field created by a changing $\vec{E}$ is called an induced magnetic field.

FIGURE 31.16 shows the close analogy between induced electric fields, governed by Faraday’s law, and induced magnetic fields, governed by the second term in the Ampère-Maxwell law. An increasing solenoid current causes an increasing magnetic field. The changing magnetic field, in turn, induces a circular electric field. The negative sign in Faraday’s law dictates that the induced electric field direction is ccw when seen looking along the magnetic field direction.

FIGURE 31.16 The close analogy between an induced electric field and an induced magnetic field.

An increasing capacitor charge causes an increasing electric field. The changing electric field, in turn, induces a circular magnetic field. But the sign of the Ampère-Maxwell law is positive, the opposite of the sign of Faraday’s law, so the induced magnetic field direction is cw when you’re looking along the electric field direction.

EXAMPLE 31.3 The fields inside a charging capacitor

A 2.0-cm-diameter parallel-plate capacitor with a 1.0 mm spacing is being charged at the rate $0.50\text{C}\text{/}\text{s}$. What is the magnetic field strength inside the capacitor at a point 0.50 cm from the axis?

MODEL The electric field inside a parallel-plate capacitor is uniform. As the capacitor is charged, the changing electric field induces a magnetic field.

VISUALIZE FIGURE 31.17 shows the fields. The induced magnetic field lines are circles concentric with the capacitor.

FIGURE 31.17 The magnetic field strength is found by integrating around a closed curve of radius r.

FIGURE 31.17 Full Alternative Text

SOLVE The electric field of a parallel-plate capacitor is $E=Q\text{/}{ϵ}_{0}A=Q\text{/}{ϵ}_0\pi R^2$. The electric flux through the circle of radius r (not the full flux of the capacitor) is

$${\text{Φ}}_{\text{e}}=\pi r^{2}E=\pi r^2{\displaystyle \frac{Q}{{ϵ}_0\pi R^2}}={\displaystyle \frac{r^2}{R^2}}{\displaystyle \frac{Q}{{ϵ}_0}}$$

Thus the Ampère-Maxwell law is

$$\oint \vec{B}\cdot d\vec{s}={ϵ}_0{\mu }_0{\displaystyle \frac{d{\text{Φ}}_{\text{e}}}{dt}}={ϵ}_0{\mu }_0{\displaystyle \frac{d}{dt}}\left({\displaystyle \frac{r^2}{R^2}}{\displaystyle \frac{Q}{{ϵ}_0}}\right)={\mu }_0{\displaystyle \frac{r^2}{R^2}}{\displaystyle \frac{dQ}{dt}}$$

The magnetic field is everywhere tangent to the circle of radius r, so the integral of $\vec{B}\cdot d\vec{s}$ around the circle is simply $BL=2\pi rB$. With this value for the line integral, the Ampère-Maxwell law becomes

$$2\pi rB={\mu }_0{\displaystyle \frac{r^2}{R^2}}{\displaystyle \frac{dQ}{dt}}$$

and thus

$$\begin{array}{ccc}\hfill B& \hfill =\hfill & {\displaystyle \frac{{\mu }_0}{2\pi }}{\displaystyle \frac{r}{R^2}}{\displaystyle \frac{dQ}{dt}}=(2.0\times {10}^{-7}\text{T}\text{m}\text{/}\text{A}){\displaystyle \frac{0.0050\text{m}}{(0.010\text{m}{)}^2}}(0.50\text{C}\text{/}\text{s})\hfill \\ \hfill & \hfill =\hfill & 5.0\times {10}^{-6}\text{T}\hfill \end{array}$$

REVIEW Charging a capacitor at 0.5 C/s requires a 0.5 A charging current. We’ve seen many previous examples in which a current-carrying wire with $I\approx 1$ A generates a nearby magnetic field of a few microtesla, so the result seems reasonable.

If a changing magnetic field can induce an electric field and a changing electric field can induce a magnetic field, what happens when both fields change simultaneously? That is the question that Maxwell was finally able to answer after he modified Ampère’s law to include the displacement current, and it is the subject to which we turn next.