# Induced-EMF and the Lorentz force?

1. Dec 14, 2014

### PhiowPhi

When a wire( that has the flow of I) is placed inside a magnetic field(B) experiences the Lorentz force and beings to accelerate in the uniform and constant magnetic field, there is no induced EMF correct? Only when it starts to exist the magnetic field there is change in flux. And there is induced EMF.

If there is still current flowing in the conductor, is there two electric fields now? One from the power source(that creates the potential difference, therefore, current flow) and one that is induced from the change of flux that reduces the applied PD( from its -PD) and by doing so, the applied current?

Also, if the wire was moving in a changing magnetic field, is the Lorentz force responsible for both motion, and induced EMF to reduce current applied to it? It confuses me that the Lorentz force simultaneously accelerates the conductor AND separates the charges to create the electric field, and the potential difference of an opposing EMF.

2. Dec 14, 2014

### vanhees71

First of all there is only one electromagnetic field, created by the charge and current densities.

Now, when a wire is moving in a static magnetic field, there are forces acting on the charges (atomic nuclei and electrons) the wire is made of. Part of the electrons, the conduction electrons, in a metal is quite free to moving around relative to the ions that are quite rigid and make up a lattice. When the motion just starts, there is a Lorentz force which, if it has a component direction of the wire, the conduction electrons start to move, leading to a build-up of corresponding opposite charges piling up on both ends of the wire, which leads to an electric field which counteracts the force on the electrons from motion in the magnetic field. Thus the motion of the electrons comes to an end as soon as the force is 0 (assuming a normal conductor not a superconductor, where the magnetic field is expelled from the conductor due to the Meissner effect).

To make this quantitative, suppose we have a homogeneous magnetic field in the $z$ direction, a wire of length $L$ extended along the $y$ direction, moving with uniform velocity in $x$ direction. In the stationary state the total force on a conduction electron must be 0, leading to
$$\vec{E}+\vec{v} \times \vec{B}=0 \; \vec{E}=-\vec{v} \times \vec{B}=-v B \vec{e}_x \times \vec{e}_z=v B \vec{e}_y.$$
Thus there is a voltage difference between the ends of the wire
$$\Delta V=v B L.$$
You can also come to this conclusion using Faraday's Law in integral form, but you have to be careful to do the time derivative correctly. It's easier using the Lorentz force to argue (as it is almost always easier to argue with the local laws and not with the integral version; that's because electromagnetism is the paradigmatic example for a relativistic field theory).

3. Dec 14, 2014

### PhiowPhi

@vanhees71, that was my initial conclusion and thank you for also reminding me of the atomic level of things shows me a bit of perspective(and what I'm missing out). But what confuses me is the use of: $$F=IL×B.$$ To calculate the kinematics of the wire since it's the force that caused motion, and at the same time the Lorentz force being the cause of opposing induced EMF? It's the force that moves the wire, and at the same time the force that makes the conduction of electrons?

My way to solve such related problems is simply use:
$$\Delta V=-v B L.$$ for induced EMF, and any other formulation using the F from the Lorentz force and I can manage. But just wanted to be sure of a few things related to the theory.