# Particles in Magnetic Fields

1. Apr 10, 2008

### rjeannier

Okay, I am in AP physics this year at my high school. We are doing E&M (Electro Magnetism) right now and I have a question that my teacher could not answer (who by the way is quiet possibly the smartest person I have ever met). Okay, I understand that a charged particle moving perpendicular through a constant magnetic field experiences a force. I also get that if the charged particle is not moving their is no force. My question is if you could create a constant magnetic field, lets say with a solenoid, and you moved that magnetic field through a stationary charged particle, would the charged particle feel a force? Similar to having a stationary plane and blowing air at it so it would take off. I feel like the particle should have a force. But if it does, what exactly is doing the moving in a magnetic field? And also, the particle has no velocity vector, so how could it be perpendicular or parallel if it is not moving? Their is a lot I am confused about here, and I have a feeling that a possible explanation would require much more than just two years of high school physics to understand.

2. Apr 10, 2008

### Astronuc

Staff Emeritus
3. Apr 10, 2008

### rjeannier

4. Apr 10, 2008

### Astronuc

Staff Emeritus
Even if a magnetic field is constant, the 'lines' of the field would move with respect to the charge. I'm trying to think of how a constant field could be move without changing the flux density - possibly a rotating solenoid, or circular permanent magnet.

5. Apr 10, 2008

### rjeannier

Yes, this is what I am asking. So if a constant magnetic field is moving and a particle is stationary, will it feel a force? I think that it would, but how can a magnetic field move? It is a field. Are their some sort of sub atomic particles or something that are doing the moving, because otherwise, how can you move something that does not physically exist.

6. Apr 11, 2008

### AstroRoyale

The two situations are completely degenerate, all motion is relative so a particle moving by a stationary field is completely the same as a constant magnetic field moving by a stationary particle. There would certainly be a force on the "stationary" particle, and it would deflect in the exact same manner

7. Apr 11, 2008

### Ben Niehoff

No, there would be NO force on the particle.

The reason is due to relativity: electromagnetic fields exhibit Lorentzian invariance; not Galilean invariance. In fact, the Lorentz-invariance of Maxwell's equations is what inspired the development of relativity in the first place, and this very example played a part.

First, consider the case of the moving particle, going through a stationary magnetic field. Obviously this particle will be deflected at right angles to its trajectory. But what does this configuration look like in the rest frame of the particle?

The error you are committing is that you are assuming that, in the rest frame of the particle, this configuration will look like a moving magnetic field passing a stationary particle. But in fact, that is not true. In reality, the electromagnetic field transforms according to a Lorentz transformation.

This is difficult to explain if you haven't studied relativity yet, but the basic idea is that the electromagnetic field is actually a single mathematical object, in the form of a 4x4 matrix (called a "tensor"), rather than two separate vector fields. It looks like this:

$$F = \left[ \begin{array}{cccc} 0 & E_x & E_y & E_z \\ -E_x & 0 & -B_z & B_y \\ -E_y & B_z & 0 & -B_x \\ -E_z & -B_y & B_x & 0 \end{array} \right]$$

(Note: You might end up with some factors of c, depending on your choice of units.)

In order to compare electromagnetic fields in different reference frames, you need to transform this matrix via a Lorentz transformation, which is basically matrix multiplication. The end result is that, when you transform from one reference frame to another, the electric and magnetic fields are allowed to mix!

If you have a pure magnetic field in one frame, you will actually have a combination of electric and magnetic fields in a moving frame. It so happens that the electric field so calculated is just the right magnitude and direction such that it produces the same perpendicular deflection force as would be observed in the case of the particle moving. This page offers more explanation:

http://en.wikipedia.org/wiki/Moving_magnet_and_conductor_problem

The main idea is that there is no such thing as a "moving magnetic field". You can have a changing magnetic field, which will produce observable effects. But if the magnetic field is constant over time, then it does not matter whether the source of the field is "moving" or not; the magnetic field is not changing.

8. Apr 11, 2008

### rjeannier

Okay, some of your explanation was slightly over my head, but I think I followed most of it. But, there has to be such a thing as a moving magnetic field because otherwise any pole sticking out from the surface of the earth is going to feel like it is moving perpendicular to Earths magnetic field. And if that is the case, then all antennas or flag poles are going to fall subject to some sort of charge separation, and that isn't the case, is it?. Now maybe I am mistaken in my assumptions, but I would have to think that the Earths magnetic field is "rotating" around the surface of the Earth at the same speed the surface of the Earth is rotating. Because then through relativity, all the poles and things jutting out of the surface of the Earth, would think they are stationary in a magnetic field. I am just trying to get a better understand of what exactly a magnetic field is. I get that its a man made concept, but it has to be something that is acting on the particles that is not just a "field". And can this something be moved? I am just really confused on the overall concept.

Last edited: Apr 11, 2008
9. Apr 11, 2008

### Ben Niehoff

What reference do you have to show that charge separation in conducting poles isn't in fact the case? That seems to be what you're hung up on.

The force on the charges in a conducting pole at the Earth's surface, due to the Earth's magnetic field, is very small: on the order of 0.03 N/C. And you realize that if any charges separate, they will be strongly attracted to each other by their electric field. The result is that an equilibrium should be reached with a very small charge separation, inversely proportional to the height squared of the antenna in question (i.e., smaller antennas will admit much smaller charge separation, due to the inverse-square law of the electric force between the separated charges).

10. Apr 11, 2008

### Ben Niehoff

As an example, consider a vertical conducting rod of 1 meter length.

First, because it is a conductor, then in equilibrium, any excess separation charge will be on the ends. Specifically, you will have a positive charge Q on the top, and a negative charge -Q on the bottom. There will be no net charge in between.

The velocity of the Earth's rotation at the equator is approximately 1000 mph. In SI units, this is about 447 m/s.

The strength of the Earth's magnetic field, measured at the surface of the Earth, is around 3 x 10^-5 T.

So, the magnetic force pushing the positive charge UP is approximately (0.013 * q) newtons. OK.

But there is an electric force pulling the positive charge back DOWN, due to its attraction to the negative charge at the bottom of the conducting rod. This force is given by:

$$F = \frac1{4\pi\epsilon_0} \frac{q^2}{d^2}$$

where d is one meter (the length of the rod). This works out to:

F = (8.99 x 10^9) * q^2 newtons

So, to find the equilibrium charge separation, we simply find the q at which these two forces balance each other:

(8.99e9) q^2 = (0.013) q

q = (0.013) / (8.99e9) = 1.49 x 10^-12 coulombs

The voltage difference due to this separation charge is on the order of 25 mV, which is too low to be accurately measured by most common voltmeters.