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Reading Wangsness's Magnetic Fields book I stumbled upon an explanation I can't understand. It goes like this: There's an alternating current solenoid with radius $a$. We know the magnetic induction is

$$\textbf{B} = \begin{cases} B_{0}\cos(\omega t+\alpha)\textbf{k}& \text{ for } r\leq a, \\ 0& \text{ for } r\geq a. \end{cases}$$

It says that due to the problem's symmetry we can expect an electric field on the xy-plane with just a $\phi$ component so

$$\oint_C\textbf{E}\cdot d\textbf{s}=\oint_C E_{\phi}\rho d\phi=2\pi\rho E_{\phi}.$$

My question is: How de we know we can expect this? Why not to expect a $\rho$ component for the field?

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4 Answers 4

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According to the Maxwell equations, in the areas where the current density is zero, the temporal derivative of the electric field equals the curl of the magnetic field (up to a constant factor). You may wish to look at the expression for a curl of a vector function in cylindrical coordinates (http://isites.harvard.edu/fs/docs/icb.topic970148.files/Spherical_coord.pdf)

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Isn't the simple answer Gauss's law? The net charge inside any imaginary cylinder co-axial with the solenoid is zero, so the net electric flux emerging through the cylinder walls is zero. But by symmetry, if there were a radial electric field component it would be the same at all points on the cylinder so there would be either a net outward flux or a net inward flux. So the radial electric field component must be zero.

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  • $\begingroup$ I've just spotted that this answer is essentially the same as the (not very well hidden) deleted answer of Hal Hollis. $\endgroup$ Dec 23, 2021 at 18:36
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Sorry for my poor english. My native language is french.

A general symmetry argument can be used:

*) If a system admits a symmetry plane (invariant by mirror symmetry), the electric field (true vector or polar vector) at a point on the plane is in the plane while the magnetic field (pseudo vector or axial vector) is orthogonal to the plane.

*) It is the opposite for an "antisymmetry" plane (system invariant by a mirror symmetry followed by a change of the sign of the charges).

In your example, the plane containing the axis of the solenoid and passing through a point M is an antisymmetry plane and therefore the electric field at point M is orthoradial.

These symmetry properties are contained in Maxwell's equations. They could be proved by using directly Jefimenko's equations as we do in statics with Coulomb's and Biot and Savart's laws.

But one can also justify them by invoking the invariance of electromagnetism under parity: One cannot distinguish right and left with an electromagnetic machine. And so, if we build a machine that is the image in a mirror of a first machine, it continues to function by remaining the image in a mirror of the first machine. The Lorentz force must therefore be transformed accordingly. enter image description here

For a system which is its own image (symmetrical), this requires that the electric field at two symmetrical points is symmetrical and that the magnetic field is "antisymmetrical" (because of the vector product).

For an "antisymmetric" system: it only becomes identical to itself after a mirror symmetry and a change of sign of the charges which leads to a change of sign of the fields. Thus, at two symmetric points, the electric field is "antisymmetric" and in particular at a point of an antisymmetric plane of symmetry, the field is normal to the plane.

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The relevant equation is

\begin{equation} \nabla \times \textbf{E} = -\frac{\partial \textbf{B}}{\partial t}. \end{equation}

$\textbf{B}$ is directed along the axis of the solenoid, i.e. the $\hat{z}$ direction, so we're going to integrate both sides over a surface which is the cross-section of the solenoid - i.e. a circle with radius $a$, on the axis of the solenoid and with normal perpendicular to the $\hat{z}$ direction

\begin{equation} \begin{split} \int \nabla \times \textbf{E} \cdot \text{d}\textbf{S}= -\frac{\partial }{\partial t} \int \textbf{B} \cdot \text{d}\textbf{S} \\ \oint_{C} \textbf{E}\cdot\text{d}\textbf{l} = \int \omega B_{0}\sin{(\omega t + \alpha)} \text{d}S \\ \int_{-\pi}^{\pi}E_{\phi}\rho\text{d}\phi = \int \omega B_{0}\sin{(\omega t + \alpha)} \text{d}S \end{split} \end{equation}

where we have used Stokes theorem on the LHS of the equation between the first two lines and $C$ is the loop bounding our circular cross section. Since this circular path is always directed along $\hat{\phi}$ the closed loop integral only picks out the $\phi$ component of the electric field.

Moreover, using Gauss' law and the fact that there is no charge inside the solenoid

\begin{equation} \oint \textbf{E} \cdot \text{d}\textbf{S} = 0 \end{equation}

where the closed surface we are integrating over is a cylinder coaxial with the solenoid, $E_{\rho} = E_{z} = 0$, since these are the only components of the field contributing to the surface integral.

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  • $\begingroup$ Just because the integral is only dependant on $E_{\phi}$ doesn't mean E only has a $\hat \phi$ component. Likewise, just because we pick a cylindrical surface, whose surface integral depends upon $\hat \rho$ , $\hat \phi$ components of the $\vec{E}$ field, doesn't mean $E_{\rho}= E_{z}= 0$. For example , a square shaped B field also has zero flux, and the components aren't 0. The fundamental reason is the geometry of the B field, not analysing the flux /line integral. The only way you can answer OP's question, is to evoke symmetry, which can be more easily done by the definition of curl $\endgroup$ Jun 8 at 17:30
  • $\begingroup$ I'm not trying to use the fact that the integral only depends on $E_{\phi}$ to justify that that is the only non-zero component, I'm simply clarifying where the equation comes from in case this was also contributing to the OPs confusing. I see your point regarding the cylindrical surface and agree that (according to the right hand screw rule) it is a symmetry argument that should be used to determine the orientation of $E$. $\endgroup$
    – Niall
    Jun 8 at 17:35

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