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13.4: Induced Electric Fields

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    4431
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    The fact that emfs are induced in circuits implies that work is being done on the conduction electrons in the wires. What can possibly be the source of this work? We know that it’s neither a battery nor a magnetic field, for a battery does not have to be present in a circuit where current is induced, and magnetic fields never do work on moving charges. The answer is that the source of the work is an electric field \(\vec{E}\) that is induced in the wires. The work done by \(\vec{E}\) in moving a unit charge completely around a circuit is the induced emf \(ε\); that is,

    \[\epsilon = \oint \vec{E} \cdot d\vec{l},\] where \(\oint\) represents the line integral around the circuit. Faraday’s law can be written in terms of the induced electric field as

    \[\oint \vec{E} \cdot d\vec{l} = - \dfrac{d\Phi_m}{dt}.\]

    There is an important distinction between the electric field induced by a changing magnetic field and the electrostatic field produced by a fixed charge distribution. Specifically, the induced electric field is nonconservative because it does net work in moving a charge over a closed path, whereas the electrostatic field is conservative and does no net work over a closed path. Hence, electric potential can be associated with the electrostatic field, but not with the induced field. The following equations represent the distinction between the two types of electric field:

    \[ \underbrace{\oint \vec{E} \cdot d\vec{l} \neq 0}_{\text{Induced Electric Field}}\]

    \[\underbrace{ \oint \vec{E} \cdot d\vec{l} = 0}_{\text{Electrostatic Electric Fields}}.\]

    Our results can be summarized by combining these equations:

    \[\epsilon = \oint \vec{E} \cdot d\vec{l} = - \dfrac{d\Phi_m}{dt}. \label{eq5}\]

    Example \(\PageIndex{1}\): Induced Electric Field in a Circular Coil

    What is the induced electric field in the circular coil of [link] (and [link]) 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 \(\vec{E}\) is tangent to the coil,

    \[\oint \vec{E} \cdot d\vec{l} = \oint E dl = 2 \pi r E.\]

    When combined with Equation \ref{eq5}, this gives

    \[E = \dfrac{\epsilon}{2\pi r}.\]

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

    \[E(t_1) = \dfrac{6.0 \space V}{2\pi \space (0.50 \space m)} = 1.9 \space V/m;\]

    \[E(t_2) = \dfrac{4.7 \space V}{2\pi \space (0.50 \space m)} = 1.5 \space V/m;\]

    \[E(t_3) = \dfrac{0.040 \space V}{2\pi \space (0.50 \space m)} = 0.013 \space V/m;\]

    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 \(dB/dt \neq 0\) 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 \(dB/dt \neq 0\) whether or not there is a conducting path present.

    These nonconservative electric fields always satisfy Equation \ref{eq5}. For example, if the circular coil of [link] were removed, an electric field in free space at \(r = 0.50 \space m\) would still be directed counterclockwise, and its magnitude would still be 1.9 V/m at \(t = 0\). 1.5 V/m at \(t = 5.0 \times 10^{-2}s\), etc. The existence of induced electric fields is certainly not restricted to wires in circuits.

    Example \(\PageIndex{1}\): Electric Field Induced by the Changing Magnetic Field of a Solenoid

    Part (a) of Figure shows a long solenoid with radius R and n turns per unit length; its current decreases with time according to \(I = I_0 e^{-\alpha t}\). 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]. (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.

    Figure A shows a side view of the long solenoid with the electrical current flowing through it. Figure B shows a cross-sectional view of the solenoid from its left end.

    Figure \(\PageIndex{1}\): (a) The current in a long solenoid is decreasing exponentially. (b) A cross-sectional view of the solenoid from its left end. The cross-section shown is near the middle of the solenoid. An electric field is induced both inside and outside the solenoid.

    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 \[B = \mu_0 nI = \mu_0 n I_0 e^{-\alpha t}.\]  Thus, the magnetic flux through a circular path whose radius r is greater than R, the solenoid radius, is \[\Phi_m = BA = \mu_0 n I_0 \pi R^2 e^{-\alpha t}.\]  The induced field \(\vec{E}\) is tangent to this path, and because of the cylindrical symmetry of the system, its magnitude is constant on the path. Hence, we have \[\left| \oint \vec{E} \cdot d\vec{l}\right| = \left|\dfrac{d\Phi_m}{dt}\right|,\] \[E(2\pi r) = \left|\dfrac{d}{dt} (\mu_0 n I_0 \pi R^2 e^{-\alpha t})\right| = \alpha \mu_0 n I_0 \pi R^2 e^{-\alpha t},\] \[E = \dfrac{\alpha \mu_0 n I_0 R^2}{2r} e^{-\alpha t} \space (r > R).\]
    2. For a path of radius r inside the solenoid, \(\Phi_m = B\pi r^2\), so \[E(2\pi) = \left|\dfrac{d}{dt} (\mu_0 n I_0 \pi r^2 e^{-\alpha t})\right| = \alpha \mu_0 n I_0 \pi r^2 e^{-\alpha t},\] and the induced field is \[ E = \dfrac{\alpha \mu_0 n I_0 r}{2} e^{-\alpha t} \space (r < R).\]
    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 \(|\vec{E}|\) increases with r inside and decreases as 1/r outside the solenoid, as shown in Figure.

    Figure is a plot of the electric field E versus distance r. Electric field is zero at the beginning, rises linearly till r equal to R, reaches sharp maximum at R, and falls of proportional to 1/r

    Figure \(\PageIndex{2}\): The electric field vs. distance r. When \(r < R\), the electric field rises linearly, whereas when \(r > R\), the electric field falls of proportional to 1/r.

    Note

    Check Your Understanding Suppose that the coil of [link] is a square rather than circular. Can Equation be used to calculate (a) the induced emf and (b) the induced electric field?

    Solution

    a. yes; b. Yes; however there is a lack of symmetry between the electric field and coil, making \(\oint \vec{E} \cdot d\vec{l}\) a more complicated relationship that can’t be simplified as shown in the example.

    Note

    Check Your Understanding What is the magnitude of the induced electric field in Example at \(t = 0\) if \(r = 6.0 \space cm\), \(R = 2.0 \space cm\), \(n = 2000\) turns per meter, \(I_0 = 2.0 \space A\), and \(\alpha = 200 \space s^{-1}\)?

    Solution

    \(3.4 \times 10^{-3} \space V/m\)

    Note

    Check Your Understanding The magnetic field shown below is confined to the cylindrical region shown and is changing with time. Identify those paths for which \(\epsilon = \oint \vec{E} \cdot d\vec{l} \neq 0\).

    Figure shows the magnetic filed confined within the cylindrical region. Area P1 partially lies in the magnetic field. Area P2 is larger that the area of magnetic field and completely includes it. Area P3 lies outside of the magnetic field. Area P4 is smaller than the area of the magnetic field and is completely included within it.

    Solution

    \(P_1, \space P_2, \space P_4\)

    Note

    Check Your Understanding A long solenoid of cross-sectional area \(5.0 \space cm^2\) is wound with 25 turns of wire per centimeter. It is placed in the middle of a closely wrapped coil of 10 turns and radius 25 cm, as shown below. (a) What is the emf induced in the coil when the current through the solenoid is decreasing at a rate \(dI/dt = -0.20 \space A/s\)? (b) What is the electric field induced in the coil?

    Figure shows a long solenoid placed in the middle of a closely wrapped coil with radius of 25 cm.

    Solution

    a. \(3.1 \times 10^{-6} V\); b. \(2.0 \times 10^{-7} \space V/m\)

    Contributors

    Paul Peter Urone (Professor Emeritus at California State University, Sacramento) and Roger Hinrichs (State University of New York, College at Oswego) with Contributing Authors: Kim Dirks (University of Auckland) and Manjula Sharma (University of Sydney). This work is licensed by OpenStax University Physics under a Creative Commons Attribution License (by 4.0).