# 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

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
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 $\vec v \times \vec B$.