Relativity, Electric Currents and Magnetic Fields

[LucasGB]

Suppose we have a current-carrying wire and an electric charge which is moving parallel to the wire with the same speed and direction as the electrons in the wire. The wire produces a magnetic field which acts upon the charge through a force. From the charge's reference frame, though, no magnetic field can be said to exist, since the electrons in the wire are at rest with respect to it. The force on the charge is thereofre justified through the following reasoning: the motion of the positive nucleus in the wire with respect to the carge causes a Lorentz contraction which increases the positive volume charge density in the wire, leaving the negative volume charge density unchanged, since the electrons are not moving in this particular reference frame. This unbalance gives rise to an electrical force which acts on the wire.

The question: what if my current does not consist of electrons in a wire, but of an electron beam in vacuum? Since there's no positive charges in any reference frame, how can I explain, from the charge's point of view the existence of the electric field?

 

[marcusl]

The test charge sees a line of electrons that appear to be at rest. It is repelled by the electrostatic Coulomb force.
EDIT: Sorry, the last part of my post didn't appear: The charge density of the electrons seen by the moving test charge is lower than that seen in the lab frame, so the Coulomb repulsion is smaller. This reduction arises, in the lab frame, from a magnetic attractive force.

 

[collinsmark]

Hello LucasGB,

Let's not stop there, but instead let's make it even simpler!

Let's just have two point charges total. Suppose the point charges are held in place, one slightly above the other, and the charges (and whatever is holding them together) are put on a train.

According to an observer on the ground, the charges not only have electric force on each other, but they also have a magnetic force when the train moves. The electric force can be described easily enough,

$$F_E = \frac{1}{4 \pi \epsilon _0} \frac{q_1 q_2}{r^2}$$

The direction of this electric force is away from the other charge (this force repels the charges).

Since the train is moving, the observer on the ground sees a magnetic force too acting on the two charges. The magnetic field created by the first charge is:

$$\vec B = \frac{\mu _0}{4 \pi}\frac{q_1(\vec v \times \hat r)}{r^2}$$

and the magnetic force on the other charge (by it moving through the field) is,

$$\vec F_M = q_2(\vec v \times \vec B)$$

Combining the above equations, and simplifying the direction a bit, we get

$$F_M = \frac{\mu_0}{4 \pi}\frac{q_1 q_2 v^2}{r^2}$$

and the direction of this magnetic force is toward the other charge (this force attracts the charges).

So the electric force tends to push the charges apart. The magnetic force tends to pull them together, but to a lesser extent. And the magnetic force increases with the velocity of the train.

So as the train moves faster, the net repulsive forces on the charges become less, at least as observed by someone at the station (standing on the ground). That means if they were released, they would accelerate away from each other at a slower rate, depending on how fast the train is moving (as measured in the frame of reference at the station).

So how fast would the train have to be moving for the magnetic force to catch up with the electric force, such that the forces were completely equal? Let's find out. We set F_E and F_M equal to each other and solve for v.

$$\frac{1}{4 \pi \epsilon _0} \frac{q_1 q_2}{r^2} = \frac{\mu_0}{4 \pi}\frac{q_1 q_2 v^2}{r^2}$$

Solving for v, we get,

$$v = \frac{1}{\sqrt{\epsilon _0 \mu_0}}$$

So if the velocity of the train approaches $$1/ \sqrt{\epsilon _0 \mu_0}$$ the magnetic force [almost] equals the electric force, and the charges don't accelerate away from one another [significantly].

It's no coincidence that this velocity, $$1/ \sqrt{\epsilon _0 \mu_0}$$, is c, the speed of light in a vacuum.

So what's really going on here? From the point of view of an observer in the station, time is moving very slowly in the moving train, and in this example, the increased magnetic force goes along with the mathematical description.

This treatment of charges might be a good first step to apply to your beam of electrons in a vacuum.

 

[Meir Achuz]

It's more complicated than that. The E and B fields have v^2 corrections that have to be included. The correction to E is the same magnitde as the v^2 B effect.
The unambiguous answer is given by finding E, B and F in the rest frame, and then Lorentz transforming everything. You will get F'=q_2(E'+vXB').

 

[LucasGB]

Thank you all for your very interesting replies. I believe the questions is answered.