## EMF in a solenoid

A colleague identified a puzzle concerning Faraday’s law. A circular loop at the center of a long solenoid with time-varying magnetic field will run a current due to the emf, which is the path integral of the (curly) non-Coulomb electric field. Put another loop beside the centered one and the puzzle is that the non-Coulomb electric field must be in the same direction in both loops where the two loops nearly touch, yet current will run in the same direction in both (clockwise or counterclockwise). Here’s a solution to the puzzle.

In the figure, the dashed circle is centered in a solenoid in which there is a time-varying spatially-uniform magnetic field that is increasing with time. By symmetry, the non-Coulomb (NC) electric field is tangent to the dashed circle as indicated by the green arrows. By “non-Coulomb” is meant that this is a curly electric field associated with a time-varying magnetic field, not a non-curly “Coulomb” electric field associated with stationary charges.

At larger radii, $E_{NC}$ increases linearly with increasing radius as indicated by the green arrows to the right of the dashed circle, as can be seen from this calculation, where k is a constant:

$2\pi rE = \pi r^2 dB/dt$

$E = kr$

The emf around the dashed circle, the path integral of the electric field, is

$\text{emf}_1 = (2\pi r_1)(kr_1)$

If we were to put a metal ring with resistance per unit length b where the dashed circle is, there would be a current in the ring of amount

$I_1 = \text{emf}_1/(2\pi r_1 b) = E_1L_1/(2\pi r_1 b) = (k r_1)(2\pi r_1)/(2\pi r_1 b) = kr_1/b$

Next consider the path marked by a blue line, which has been constructed to enclose the same amount of area as is enclosed by the dashed circle:

$\pi r_1^2 = \dfrac{1}{4}(\pi r_2^2 - \pi r_1^2)$

$\dfrac{5}{4}r_1^2 = \dfrac{1}{4}r_2^2$

$r_2 = \sqrt{5}r_1$

Because the two paths enclose the same area, with $dB/dt$ the same everywhere, the emf must be the same around both paths. Moving counterclockwise from the lower left corner of the second path, we can calculate the path integral of the non-Coulomb electric field, and we do indeed find the same emf as that for the circular path:

$\text{emf}_2 = 0 + (kr_2)(2\pi r_2)/4 + 0 + (-kr_1)(2\pi r_1)/4$

$\text{emf}_2 = 2\pi(kr_2^2 - kr_1^2)/4$

$\text{emf}_2 = 2\pi(5kr_1^2 - kr_1^2)/4 = 2\pi kr_1^2$

$\text{emf}_2 = (2\pi r_1)(kr_1) = \text{emf}_1$

If we place a metal wire along the outer path, with the same resistance b per unit length, the current $I_2$ will be less than $I_1$ because the path $L_2$ is longer than the path $L_1$ ($2\pi r_1$):

$L_2 = 2(r_2-r_1) + \dfrac{2\pi r_2}{4} + \dfrac{2\pi r_1}{4} = 1.20L_1$

With the same emf along the longer length of wire, $I_2 = I_1/1.20$.

An important difference between the two configurations of wire is the distribution of polarization charges on the surface of the wire. In the circular case, electrons are accelerated inward along their circular path, so the wire must be polarized transverse to the wire, with the outer part of the wire charged negatively and the inner part charged positively, to provide an outward-pointing Coulomb electric field that accelerates the electrons inward. This transverse polarization is extremely small because in a good metal conductor the drift speed (and therefore the centripetal acceleration $v^2/r$) is extremely small.

The situation is much more complex when a wire follows the outer path. Conservation of charge in the steady state, with a wire of constant cross section and uniform resistivity, means that the current $I_2$ must be the same all along the path. This in turn means that the net electric field, non-Coulomb field plus Coulomb field due to surface charges, must have the same magnitude all along the path, as is indicated by the orange arrows in the diagram. These arrows are drawn slightly shorter than the arrows drawn on the circular path, because the emf is the same but the wire length is 1.20 as long.

The non-Coulomb electric field $E_\text{NC}$ polarizes the metal, making the upper left section of the wire positively charged and the lower right section negatively charged. These polarization charges are the source of a Coulomb field $E_\text{C}$ which, when added to the non-Coulomb field, makes a net electric field that has constant magnitude around the path, corresponding to the constant current around the path. The Coulomb field can be visualized qualitatively by thinking about what Coulomb field is required to add to the non-Coulomb field to give the indicated net field $E_\text{net}$.

Note that the path integral of the Coulomb field $E_\text{C}$ will be zero, because the electric field made by charges has no curl. That part of the net field that is the non-Coulomb field is solely responsible for the non-zero path integral that is the emf.

For clarity in the diagram the arrows representing electric field are drawn just outside the paths, but in the presence of a wire they represent the electric field inside the wire; surface charges will contribute to non-zero Coulomb field outside the wire with components perpendicular to the wire.

Bruce Sherwood

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