Worked exercise — Varying B field generates circulation of E

Problem

In a region of space there is a magnetic field, uniform in direction, perpendicular to a flat surface $S = 2\;\text{m}^2$, whose magnitude grows in time according to $\abs{\vv B(t)} = 2t^2$ tesla (with $t$ in seconds). Calculate, at the instant $t = 3\;\text{s}$, the circulation of the electric field along the closed boundary $\gamma$ of $S$.

Solution

Setup. The Faraday-Neumann law (Maxwell’s third equation in integral form) relates the circulation of the electric field along a closed line $\gamma$ to the time variation of the magnetic flux linked with the surface $S$ bounded by $\gamma$:

$\Gamma_E = \oint_\gamma \vv E \cdot \dd\vv\ell = -\ddt{\Phi_B}$

Magnetic flux. Since $\vv B$ is perpendicular to the surface and uniform over it, the flux is simply magnitude times area:

$\Phi_B(t) = \abs{\vv B}\,S = (2t^2)\cdot 2 = 4t^2 \;\text{Wb}$

Derivative of the flux. We differentiate with respect to time:

$\ddt{\Phi_B} = \ddt{}\left(4t^2\right) = 8t \;\text{V}$

Evaluation at $t = 3\;\text{s}$.

$\left.\ddt{\Phi_B}\right|_{t=3} = 8\cdot 3 = 24 \;\text{V}$

Circulation. By Faraday’s law:

$\ev{\Gamma_E = -\ddt{\Phi_B} = -24\;\text{V}}$

The minus sign expresses Lenz’s law: the induced electric field opposes the increase in flux, i.e. it is oriented so as to generate (if there were a conductor) a current that counteracts the growth of $\vv B$.

Variant. If instead the field decreased exponentially, $\abs{\vv B(t)} = B_0\,e^{-t/\tau}$, then $\Phi_B = B_0 S\,e^{-t/\tau}$ and the induced electromotive force would be

$\text{e.m.f.} = -\ddt{\Phi_B} = \frac{B_0 S}{\tau}\,e^{-t/\tau}$

positive and gradually decreasing: the slowly dying field generates a circulation of the same sign as the residual flux.

Links

Topics: Electromagnetic waves Concepts: Faraday-Neumann-Lenz law · Magnetic flux