Induced Electric Field in a Circular Coil

What is the induced electric field in the circular coil of Example 13.2 (and Figure 13.9) at the three times indicated?

Strategy

Using cylindrical symmetry, the electric field integral simplifies into the electric field times the circumference of a circle. Since we already know the induced emf, we can connect these two expressions by Faraday’s law to solve for the induced electric field.

Solution

The induced electric field in the coil is constant in magnitude over the cylindrical surface, similar to how Ampere’s law problems with cylinders are solved. Since $\overrightarrow{E}$ is tangent to the coil,

When combined with Equation 13.12, this gives

The direction of $\epsilon$ is counterclockwise, and $\overrightarrow{E}$ circulates in the same direction around the coil. The values of E are

Significance

When the magnetic flux through a circuit changes, a nonconservative electric field is induced, which drives current through the circuit. But what happens if $dB\text{/}dt\ne 0$ in free space where there isn’t a conducting path? The answer is that this case can be treated as if a conducting path were present; that is, nonconservative electric fields are induced wherever $dB\text{/}dt\ne 0,$ whether or not there is a conducting path present.

These nonconservative electric fields always satisfy Equation 13.12. For example, if the circular coil of Figure 13.9 were removed, an electric field in free space at $r=0.50\text{m}$ would still be directed counterclockwise, and its magnitude would still be 1.9 V/m at $t=0$, 1.5 V/m at $t=5.0\times 10^{\text{−}2}\text{s},$ etc. The existence of induced electric fields is certainly not restricted to wires in circuits.

Electric Field Induced by the Changing Magnetic Field of a Solenoid

Part (a) of Figure 13.18 shows a long solenoid with radius R and n turns per unit length; its current decreases with time according to $I=I_0{e}^{\text{−}\alpha t}.$ What is the magnitude of the induced electric field at a point a distance r from the central axis of the solenoid (a) when $r>R$ and (b) when $r<R$ [see part (b) of Figure 13.18]. (c) What is the direction of the induced field at both locations? Assume that the infinite-solenoid approximation is valid throughout the regions of interest.

Strategy

Using the formula for the magnetic field inside an infinite solenoid and Faraday’s law, we calculate the induced emf. Since we have cylindrical symmetry, the electric field integral reduces to the electric field times the circumference of the integration path. Then we solve for the electric field.

Solution

1. The magnetic field is confined to the interior of the solenoid where
   $$B={\mu }_{0}nI={\mu }_{0}n{I}_0{e}^{\text{−}\alpha t}.$$
   Thus, the magnetic flux through a circular path whose radius r is greater than R, the solenoid radius, is
   $${\text{Φ}}_{\text{m}}=BA={\mu }_{0}n{I}_0\pi R^2{e}^{\text{−}\alpha t}.$$
   The induced field $\overrightarrow{E}$ is tangent to this path, and because of the cylindrical symmetry of the system, its magnitude is constant on the path. Hence, we have
   $$\begin{array}{ccc}\hfill \left|{\displaystyle \oint \overrightarrow{E}·d\overrightarrow{l}}\right|& =\hfill & \left|\frac{d{\text{Φ}}_{\text{m}}}{dt}\right|,\hfill \\ \\ \\ \hfill E(2\pi r)& =\hfill & \left|\frac{d}{dt}({\mu }_{0}n{I}_0\pi R^2{e}^{\text{−}\alpha t})\right|=\alpha {\mu }_{0}n{I}_0\pi R^2{e}^{\text{−}\alpha t},\hfill \\ \\ \\ \hfill E& =\hfill & \frac{\alpha {\mu }_{0}n{I}_0{R}^2}{2r}{e}^{\text{−}\alpha t}(r>R).\hfill \end{array}$$
2. For a path of radius r inside the solenoid, ${\text{Φ}}_{\text{m}}=B\pi r^2,$ so
   $$E(2\pi r)=\left|\frac{d}{dt}({\mu }_{0}n{I}_0\pi r^2{e}^{\text{−}\alpha t})\right|=\alpha {\mu }_{0}n{I}_0\pi r^2{e}^{\text{−}\alpha t},$$
   and the induced field is
   $$E=\frac{\alpha {\mu }_{0}n{I}_{0}r}{2}{e}^{\text{−}\alpha t}(r<R).$$
3. The magnetic field points into the page as shown in part (b) and is decreasing. If either of the circular paths were occupied by conducting rings, the currents induced in them would circulate as shown, in conformity with Lenz’s law. The induced electric field must be so directed as well.

Significance

In part (b), note that $\left|\overrightarrow{E}\right|$ increases with r inside and decreases as 1/r outside the solenoid, as shown in Figure 13.19.