Recent content by 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...

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 ?

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.

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...

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.

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...

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...

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...

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...