# Lorentz force of a charged particle between moving parallel conductors

1. Jun 13, 2011

### gilbertopin

Let us consider a fixed Carthesian reference frame (x y z) with a charged (+) particle locate in its origin. Moreover, let us consider two parallel conductors of infinite lenght, carrying the same current I (in the same direction y). The two conductors move at constant velocity from the origin, starting from time t=0 (the two conductors then occupy the loci x=Vt and x = -Vt, for all t>0) . My question is: what is the correct behavior among the two listed below?
1) Being the induction B field constantly B = 0 incorrespondence of the charge, then no Force is exerted on the particle.
2) The effect of the two moving conductors sum at Electric field level, such as the charge in the origin is subjected to the force F = 2 q cross(-V,B(t)), being B(t) the field due to a single conductor at time t and -V the velocity of the particle relative to the induction field.

2. Jun 13, 2011

### chrisbaird

I believe (1) is the right answer. If the plates are really infinite and carry the same current in the same direction, then the magnetic field is zero at all points between the plates and for all times.

3. Jun 14, 2011

### gilbertopin

Well, In my opinion the effects of the two conductors on the charge sum at force level. That is, even if the magnetic field is null in correspondence of the charge at any time instant, we have an Electric field on the charge (and a force as a direct consequence) due to the displacement of the charges in the conductors. Indeed, the currents in the conductors can be only to charges that move across the space, and the universe acts as a huge condenser with plates at infinite distance from the charge.

4. Jun 14, 2011

### Staff: Mentor

Are the conductors wires carrying current or plates carrying charge? Also, is the charge at the origin moving?

5. Jun 14, 2011

### gilbertopin

a) the conductors are parallel wires carrying currents, that flow in the same direction in both the wires. The wires move away from the origin at constant velocity in opposite directions. (The current flows in direction normal to the velocity of the conductors).
b) the charge in the origin has zero velocity at time t = 0. Assume that at time t =0 the two wires are located at a non negligible, but finite distance from the origin (symmetrically). With respect to my first post, let us modify the loci as
x = Vt+L and x = -Vt-L, with L finite and arbitrary. This modification is due to the necessity to avoid the singularity that occurs when the distance between the wires and the charge becomes 0.

Last edited: Jun 14, 2011
6. Jun 14, 2011

### gilbertopin

I think that even if the conductors are taken as plates carrying charges, for the sake of this discussion it would make no difference. Wires carrying currents can be viewed as charges of opposite signs moving in opposite direction. Assume that that the positive ad negative charges are distributed evenly such that the static Electric field in the space is only due to the charge in origin.

7. Jun 14, 2011

### Staff: Mentor

The way to solve this is conceptually straight forward, but rather tedious. Start with the known B field for an infinite wire, apply the Lorentz transform to the coordinates and the fields to get the field of a wire moving at v, add two of those together with opposite v to get the total field, evaluate at the origin, and apply the Lorentz force law.

8. Jun 15, 2011

### chrisbaird

Conceptually, a wire current moving away from a fixed charge is equivalent to a charge moving away from a fixed wire current. The mechanisms are different (or at least appear to be), but he resultant force is the same. So for this problem, consider a line current in the y direction fixed at the origin while a positive electrics charge moves with constant velocity v in the x direction. It experiences a force in the y direction. With a line current fixed at the origin and a charge moving in the x-direction, it also experiences a force in the y direction. So they add together and the charge experiences a net force in the y direction, even though the magnetic field is zero. The changing magnetic field induces an electric field which exerts a force on the stationary charge.

9. Jun 15, 2011

### Staff: Mentor

Yes, that is the key.

10. Jun 15, 2011

### gilbertopin

Yes, we came to the same conclusion at the end: the effects sum at force level, not at B field level. So, basically, the net effect is that the charge experiences a force which is double than the one experienced in presence of one wire current.

11. Jun 15, 2011

### Staff: Mentor

They sum at any level. Maxwells equations and the Lorentz force law are all linear. You can sum it at the source level, at the field level, or at the force level, you will get the same answer. That is exactly why we like linear equations.

12. Jun 16, 2011

### gilbertopin

Well, we agree in the result, at least, but your conclusion is not confirmed by the result. Indeed if you sum up the magnetic field contribution of the two wires in the origin, you get B=0, that is no field in the origin. Since you neither have electric field there, in this case one should conclude that the particle in the origin is not subjected to any force.

Conversely, if you sum up the Lorentz forces computed by considering one wire at a time, then you have a force on the particle.

Since we agreed on the second answer, this means that, to make it right, you must sum up the effects at a force level. The fact is that the Maxwell's equations are linear only if you consider all the terms which make it conservative, including the displacement effets (in the classical notation the displacement is seen as a variation in time of the electric field). The problem here consists in recognizing that the displacement current is not due to a time-variation of electric field, but in a change of the momentum between the 'probe' charge in the origin and the moving charges that generate the magnetic field.

13. Jun 16, 2011

### Staff: Mentor

This is incorrect, there is an electric field there. The electric field is the source of the force in this scenario. And even if there were a magnetic field there it would not be relevant since the charge is not moving.

Obviously, you have to avoid making arithmetic or analysis mistakes like this, but if you do all the physics and math correctly then it doesn't matter what level you add it at. You can add the sources, fields, or forces, and you will get the same answer. That is guaranteed by the fact that Maxwell's equations and the Lorentz force law are linear.

14. Jun 17, 2011

### gilbertopin

If you compute the electric field as E = F/q, then for sure you get it, because the charge is ubjected to force. But it's not due to Coulomb's attraction/repulsion, that's what I mean.

15. Jun 17, 2011

### Staff: Mentor

The E field is due to Faraday's law of induction. Are you familiar with Maxwell's equations? We mentioned them above, but this response is confusing if you are familiar with them. I would be glad to explain in detail.

EDIT: actually, it is Ampere's law, not Faraday's law, see below

Last edited: Jun 18, 2011
16. Jun 17, 2011

### gilbertopin

Yes, I'm familiar with them. For the given example, I was not able to make them work, i.e., to achieve the same result obtained by applying the Lorentz force equation for the particle. Can you please explain where the Electric field at the origin (on the particle) takes origin ?

17. Jun 17, 2011

### Staff: Mentor

OK, for convenience I will assume units where the permittivity and permeability and c are all equal to 1, and the current is equal to -2 pi. Then the four-potential is known to be:
$$A'_0=\left(0,0,0,ln(x'^2+y'^2)\right)$$

Boosting that into a frame moving at v gives
$$A_v=\left(0,0,0,ln\left(\frac{(tv+x)^2}{1-v^2}+y^2\right)\right)$$

So the potential from 2 wires is given by
$$A=A_v+A_{-v}$$

Taking the curl of the spacelike part of A and evaluating at x=y=0 gives a magnetic field of zero, but taking the time derivative of the spacelike part of A and evaluating at x=y=0 gives an electric field of
$$E=\left( 0,0,-\frac{4}{t} \right)$$

So the electric field at the origin exists because the vector potential is time varying, and the vector potential is time varying because the wires are moving.

Last edited: Jun 18, 2011
18. Jun 18, 2011

### Staff: Mentor

Looking at the solution in more detail it turns out that this is incorrect. The E field at the origin is due to Ampere's law, not Faraday's. Since the B field at the origin is always 0 the time derivative is also always 0. But the B field has a non-zero curl at the origin.

19. Jun 18, 2011

### gilbertopin

Perfect, so the solution consists in summing up the two vector potentials to obtain the total vector potential given by the two moving wires. Then, being the scalar potential null, the derivative of the vector potential suffices in determining the electric field at the origin. The point is that the vector potential changes in the origin but its curl (the B field) remains always the same and zero. Thanks.

20. Jun 18, 2011

### Staff: Mentor

Yes, exactly. You can do the same thing with the formulation of Maxwell's equations in terms of fields, but I have a much easier time working with potentials.