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Example A is a conventional current conducting wire. Velocity of a positive charge is vdrift. Both velocity and current pointing down. Example B has similar setup but an observer is moving the same speed as the current. In example B's frame of reference I=0 and v=0. How does example A and B have the same direction of force on the positive charge?

2006-08-07 16:04:43 · 2 answers · asked by Anonymous in Science & Mathematics Physics

2 answers

Good question. Notice that current is a movement of the charges against an oppositely charged background. So if you travel with the "moving" charge, the "stationary" opposite charge will appear to be moving in the opposite direction. Which means a net "current" in the same direction. So everything in the right-hand rule points the same way.

Edit: Special Relativity isn't necessary for this one. This works even for low speeds, when Lorentz contraction is negligible. The point is that there is a magnetic field, moving with respect to the charge rather than a charge moving with respect to a magnetic field.

The dude below is right only to the extent that the magnetic field is mixed with an electric field via relativistic effects. But the effect you are asking about happens even for ordinary, human -scale speeds

2006-08-07 16:22:02 · answer #1 · answered by Benjamin N 4 · 0 0

Give me a minute and I'll lay it out for you. There's a subtlty the first dude missed.

Start with a wire that is overall neutral. There are electrons which give you the current but there are lattice atoms with positive charges that balance out the net charge of the wire. That is for every electron "flowing" in the wire, there is a lattice atom with a positive charge to keep the net charge of the wire zero.

Now if you stick a charged particle (call it P) outide the wire that moves with speed Vdrift along with the electrons. It will experience a force because it will be a moving charged particle in a magnetic field. What I'm guessing you're trying to ask is that if you looked at this from P's viewpoint? A viewpoint where there is no effects on it due to the magnetic field.

If you are in the same reference frame as P, then P is at rest! What the person who responded to this question first said is true, you'll still see a current but one going in the opposite direction due to the lattice atoms. However, this is irrelevant because magnetic fields only act on moving particles. The electric forces balance out (because our wire is overall neutral) and it experiences no magnetic force because it isn't moving. So what gives?

Here is one way around this. That is that the first of those assumptions is FALSE. The wire is NOT neutral in this frame. It's a subtlety that arises from Relativity. When we talked about the wire in what you called Example A, the lattice atoms were at rest and the electrons were moving. When we look at Example B the electrons are at rest and the lattice atoms are moving. Moving things incur a length contraction (this is the corresponding space side to the time side of that "moving clocks run slow" stuff you might have heard about).

So in set up A, our electrons were closer together (because they were moving and the distance between them was contracted) and the latice atoms were farther apart than they will be for set up B. If the charge densities balanced out in A, then they will not for B since here the electrons are farther apart and the lattice atoms are closer together creating a net positive charge.

That is where the force comes from.

Edit / Response: What I described above was meant to illustrate an idea ast to "why" the particle still experiences a force. I did it to avoid a more technical argument but I will give one now.

Benjamin, a moving magnetic field by itself doesn't justify your answer, and it isn't accurate to say that a moving magnetic field can act on a stationary charge because they ONLY act on moving charges. To help you see this, by your reasoning the magnetic field in Example A would be moving with respect to a stationary charge and thus should act on it. This obviously is not the case though.

In a given reference frame, magnetic fields only act on moving particles. What's happening here is that a given electro-magnetic field isn't comprised of fixed electrical and magnetic components. They are velocity dependent. That the "magnetic" field is moving isn't what resolves this, it is that the magnetic field is no longer a purely magnetic field, but also has electric components to it when it is looked at from P's frame, and it is the electrical components which act on P. You seem to be dismissing this, but it is NOT something you can dismiss when talking about situations like these regardless of the velocities they take place under.

I'll copy and paste the following from wikipedia

"A magnetic field is the relativistic part of an electric field, as Einstein explained in 1905. When an electric charge is moving from the perspective of an observer, the electric field of this charge due to space contraction is no longer seen by the observer as spherically symmetric due to non-radial time dilation, and it must be computed using the Lorentz transformations. One of the products of these transformations is the part of the electric field which only acts on moving charges - and we call it the "magnetic field"."
- http://en.wikipedia.org/wiki/Magnetic_field

The fields are mixed because of Lorentz contraction. What I described above was an abstraction to attatch these ideas to physical things, but the fact remains that relativisitic effects are not negligible in these cases. They are the very thing that links the electric and magnetic fields. That which acts on stationary charges and that which acts on moving charges.

Relativity is the reason why they are related, but the equations relating the velocity dependence of electro-magnetic fields came before Einstein's time. If you want to just accept that the fields have different components for different reference frames, you can. People did for quite a while before Al rolled along. That doesn't change the fact that magnetic fields don't act on stationary charges.

I hope that cleared things up.

2006-08-07 16:28:34 · answer #2 · answered by Anonymous · 0 0

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