# Moving conductors in magnetic fields: is there electric field or not?

this is my first question on PhysicsSE (I'm already an user of MathSE).

I'm a mathematics students trying to understand Faraday's law, that is

$$\varepsilon= -\frac{d \Phi_B}{dt}$$

where $\varepsilon$ means electromotive force and

$$\Phi_B=\iint \mathbf{B}\cdot d\mathbf{S}$$

means flux of magnetic field. As my textbook points out, there is an interpretation problem here: if the change in magnetic flux is due to movement of the conductor, then free charges in it are subject to Lorentz force, which then causes a current. On the contrary, if the conductor holds steady in a changing magnetic field, the induced current must be explained in terms of an electric field $\mathbf{E}$, described by the equations

$$\begin{cases} \nabla \cdot \mathbf{E}=\frac{\rho}{\varepsilon_0} \\ \nabla \times \mathbf{E} = - \frac{\partial \mathbf{B}}{\partial t}\end{cases}.$$

Question Do those equations hold if we have a moving conductor in a stationary magnetic field? I guess not: this would mean $\mathbf{E}=\mathbf{0}$. How to solve this?

• Maxwells equation always hold. – Fabian May 14 '11 at 13:10

## 1 Answer

Dear dissonance, as discussed e.g. in these two questions

the laws of electrodynamics in moving materials may be a bit subtle (but they may be determined). However, whenever the motion is uniform, it's straightforward to transform Maxwell's equations back to the rest frame of the moving objects.

In particular, a conductor will naturally have $\vec E =0$ in its rest frame while $\vec B$ is arbitrary. However, these two propositions get modified in a frame that is moving because the values of $\vec E,\vec B$ have to be transformed and mixed into one another if one changes the inertial frame. See

http://en.wikipedia.org/wiki/Classical_electromagnetism_and_special_relativity#Joules-Bernoulli_equation_for_fields_and_forces

Approximately, neglecting the terms of order $(v/c)^2$, we have $$\vec E' = \vec E + \vec v \times \vec B$$ $$\vec B' = \vec B - \frac{\vec v}{c^2} \times \vec E$$ Note that even if $\vec E=0$, it doesn't mean that the value $\vec E'$ in the moving frame is zero. Instead, it will be approximately $\vec v \times \vec B$.

In the case of Faraday's effect, this electric field will have all the usual consequences and it will move the electrons just like you expect, proving that changing magnetic fields do have an impact on the flows of electrons, whether you view the situation from the viewpoint of the wires or the magnets.