Magnetic Forces Part 3

Magnetic Forces: Part III, Motion Perpendicular to Current (via 4-potential)

Now consider the same current carrying wire as in the previous section, but this time with the test charge moving towards the wire.  Once again we will assume a wire aligned with the y axis that crosses the X axis at x = x0, and a test charge at the coordinate origin (i.e. a distance r= x0 away from the wire.  This time the test charge is moving towards the wire with speed beta_OT in the +x direction.

The wire and its currents are the same as for the last section, so the 4-potential in the observers frame is again

Transforming this to the test charge frame is a boost in the x direction.  Since the charge line is perpendicular to the direction of boost, the vector potential is only multiplied by  gamma_OT, which gives

From this we calculate the force again as

Now we need to be careful.  Since the charge is moving towards the wire, the vector potential is changing with time, so we get:

We need to be careful here too to recognize that x and dx are in the test charge frame coordinates, and that we will need to transform these to the observers frame in the next step.  Likewise the time dt is in this frame, and t will need to be transformed as well.  It turns out that the dx factors cancel out, however, and we can just do the usual transformation of the force to get the force in the observers frame:

This can be generalized for all components of the velocity and the vector potential, when the test charge is moving perpendicular to the vector potentials:

Total Magnetic Force Around a Wire

The results of the last two sections can be combined to give the total force due to a particle moving in a magnetic field.  We get:

Which can now be recognized as

OK, maybe that isn't obvious, but if you work it out, you will find that these are the same thing.

In terms of the magnetic fields this is: