# UY1: Induced Electric Field

Consider a long, thin solenoid with cross-sectional area A and n turns per unit length, which is encircled at its center by a circular conducting loop. The current flowing through the wires of the solenoid is  I and is changing at a rate of $\frac{dI}{dt}$.

The magnetic field at the center of the solenoid is given by:

$$B = \mu_{0} n I$$

The magnetic flux through the loop is given by:

\begin{aligned} \Phi_{B} &= BA \\ &= \mu_{0} nIA \end{aligned}

The induced e.m.f. will be:

\begin{aligned} \epsilon &=-\frac{d\Phi_{B}}{dt} \\ &=-\mu_{0}nA \frac{dI}{dt} \end{aligned} The changing magnetic flux causes an induced electric field in the conductor. (which drives the electron around the loop.) The electric field in the loop is not conservative:

\begin{aligned} \oint \vec{E}.d\vec{l} &= \epsilon \\ \oint \vec{E}.d\vec{l} &=-\frac{d\Phi_{B}}{dt} \end{aligned}

A varying magnetic field gives rise to an induced electric field.

The electric field will be given by: (for this example)

$$E = \frac{1}{2 \pi r} \left| \frac{d \Phi_{B}}{dt} \right|$$

Note: The line integral must be carried out over a stationary closed path.

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