# Does electric field cause changing magnetic field also?

According to Faraday's law, changing magnetic field causes eletric field (imprecise wording, but generally accurate). My question is, can Faraday's law be interpreted in opposite way - that is, electric field causing changing magnetic field? Equation itself doesn't speak for the direction, so I ask.

Yes. You can see this more clearly when working with the differential form of Maxwell's equations. You can consider the simple case of a wire of some radius with current flowing through it. If there's a small gap in the wire the current will accumulate some surface charge density at the ends of the wire at the gap. This creates a time varying electric field, similar to a parallel plate capacitor within the gap and it turns out this will create a magnetic field circulating around the wire axis, even at the gap, as if the wire was gapless with some current flowing through it everywhere.

So time varying electric fields act like currents and can create magnetic fields, i.e. act as sources for magnetic fields just as current densities do. This particular type of current is called a displacement current and was discovered by Maxwell to be a necessary modification to Ampere's law.

• sorry for the dumb question, but what do you mean by " If there's a small gap in the wire"? Do you mean for instance "if some copper is missing inside a copper wire", or if the wire is broken in 2 but the gap between them is tiny? Commented Nov 11, 2019 at 20:04

The correct interpretation in the opposite way is that a "changing" electric field produces a magnetic field. Also the change should be with respect to time. An example would be a current carrying wire. A current carrying wire produces a magnetic field around it, given by Biot-Savart's law.

The most general concept is that a moving charge creates a magnetic field in the region around it. A moving charge means that Electric field "changes" with time.

The magnetic field due to a moving charge is given by:

$\vec B=\frac{\mu_0}{4\pi r^3} q(\vec v \times \vec r )$

where $\vec r$ is the position of the moving charge w.r.t. the frame of reference.

Today's high speed circuits and measuring devices have allowed us to understand that you cannot separate out an electric from a magnetic field. One does not cause the other. When we have a changing $dE/dt$ we also have, by definition a changing $dB/dt$. Another way of saying this is that all electrical energy is a summation of $E\times H$ (TEM, Transverse Electromagnetic fields) We know this as the Poynting vector. So we can talk about $d(E\times H)/dt$.

Hence we will observe a $dB/dt$ if there is a $dE/dt$. Many people state equations like this $dE/dx = - dB/dt$. This may appear to be correct mathematically, but is not accurate in reality. A digital step travels at the speed of light in a medium regardless of the steady state binary 1 signal. A step of 0111100000 etc had a steady state $E$ field there is still a $B$ field associated with this signal.