# Magnetic Field and Electric Feild Confused

So I'm still confused on how a change in magnetic field could induce an electric field and vise versa.

When I got introduced to the idea of the induced electric field we looked at a circuit with one side being a bar cutting through the magnetic field generating an emf.

Now for this bar that's moving through the magnetic field, the charges within the bar will feel the magnetic force which in return creates a potential difference within the rod. From this potential difference THEN creates an electric field within the circuit.

But then later in class the teacher tells us that we can remove the circuit and with a change in magnetic flux there is an induced electric field. I don't understand how when there is no charge present there can be an electric field created.

Am I getting a wrong understanding using the bar situation?? If anyone can explain it better please do!

• If you want to know absolutely , then you have to consider Special Relativity – user182687 Mar 24 '18 at 10:17

In the situation with the bar moving on wire rails, when we describe the situation in the frame of the rails, electric field of substantial magnitude is present only in and near the resistor. In the wires and in the rod, electric field is negligible, since they are very good conductors.

If you're wondering how are the charge carriers forced to move in the bar without electric field in it, the answer is, they are forced to do so by the joined action of the magnetic forces and forces of constraint that keep the carriersinside the wires. Magnetic force deflects the carriers trajectories, they impact on the bar walls (or lattice of nuclei forming the bar) and since they cannot get out of the bar, the bar acts back and accelerates the carriers which leads to electric current. The forces of constraint acting on the charge carriers are believed in principle to be also of electromagnetic nature, but they are very localized so macroscopic electric field will not be measurable near the bar. Magnitude of emf for the circuit

$$|emf| = qvBL$$

can be expressed as

$$|emf| = |d\Phi/dt|$$ where $\Phi$ is magnetic flux through the circuit.

What your teacher is talking about when he says "remove the circuit" is an entirely different physical situation, where the bar is not present at all, or is at rest. Then, if the magnetic field itself is made to change (for example, the magnet is made to approach the bar) in time, then there will be vortex-like electric field everywhere including the bar and the space around it. Then, if we calculate work of this electric field per unit charge along the path $s$ where the circuit was, we can express it as $d\Phi/dt$:

$$\left | \int \mathbf E \cdot d\mathbf s \right | = \left| \frac{d\Phi}{dt}\right|.$$ Now, if there was a wire where the path $s$ goes, this would result in emf that has the same value.

As you can see, the two situations are different and the quantity calculated is also of different kind. The common point your teacher was talking about is the result that whether the magnetic flux changes by change of geometry or change of magnetic field, in both cases the emf is given by rate of change of magnetic flux. But the mechanism of how this happens is entirely different.

• Thank you for the answer. However I was wondering why the electric field is created and how when there is no charge present. – Nick Yarn Mar 24 '18 at 18:29
• In the case where the magnet is moving, the induced electric field around it is not like electrostatic field connected to charged particles according to the Coulomb law. Its field lines turn around and may close themselves like vortex, they do not end in the electric charges. – Ján Lalinský Mar 25 '18 at 0:42
• So this is just a property of the fields? Where electric fields can be created without any presents of charge by a change in magnetic field. And this is how electromagnetic waves travel. – Nick Yarn Mar 25 '18 at 22:59
• Not exactly. The electric and magnetic field due to moving magnet is still localized to the neighborhood of the magnet (the field intensity decays with distance as $1/r^2$, while for EM waves, it decays much more slowly, as $1/r$). There are electric charges and (magnetization) electric currents in the magnet, so its EM fields cannot be separated from those charges and currents. – Ján Lalinský Mar 25 '18 at 23:15