A Relativity: Test charge moving perpendicular to a current carrying wire exhibits a force

[Glare]

TL;DR Summary

Test charge moving perpendicular to a current carrying wire exhibit a force. What is the relativistic explanation in the charge rest frame?

Given a test charge moving along a current carrying wire, I'm familiar with the relativity explanation of the magnetic field at the rest frame of the wire and the electric field In the moving particle frame (where the wire is electrically charged).
My question is how relativity kicks in in the case where the test charge is moving perpendicular to the wire?
Can the force on the particle be explained by some charge distribution over the wire at the particle frame?

 

[Ibix]

Of course. Just transform the magnetic field and the current density and see what you get

 

[Glare]

Yes, but can we reach it only by using length contraction in similar way to how Einstein derived it for a charge moving along the wire?

 

[Ibix]

As far as I'm aware the explanation with different length contraction effects for electrons and protons is a post hoc description by Purcell, not a derivation by Einstein. Unless you have some other reference in mind?

Given that, I'd transform the 4-current density and see what the result looks like. Then I'd try to find an "intuitive" explanation for that.

 

[TSny]

Here's an attempt at a qualitative explanation.

Suppose we have a very long, straight wire with current. Imagine the current is due to motion of both positive and negative charge carriers.

The positive charges are shown in red and move upward. The negative charges are blue and move downward at the same speed. The charge density is the same for the positive and negative charge. The net current is upward. The magnetic field in the region to the right of the wire points into the page. So, a positive point charge that moves away from the wire toward the right will experience a magnetic force upward:

If we go to the rest frame of the moving charge $q$, the upward force on $q$ must be due to an electric field. We want to see how this happens. In the rest frame of $q$, the wire moves toward the left. So, the positive charges in the wire move upward and to the left. The negative charges in the wire move downward and to left. This is shown in the figure below. In the small region of the wire marked $A$, the red arrow indicates the velocity of the positive charge carriers, the blue arrow denotes the velocity of the negative carriers.

A uniformly moving point charge has a radial field that is stronger in directions perpendicular to the velocity direction. This relativistic effect is very small for non-relativistic speeds, but it's important here (I think).

The pattern of the field lines for the carriers in region $A$ are shown schematically below:

Finally, consider the diagram below which shows symmetrically placed regions $A$ and $B$ of the wire. The electric fields produced by the carriers in $A$ and $B$ at the location of the charge $q$ are shown.

The red arrow $E_A$ denotes the field at $q$ produced by a positive carrier in region $A$. The blue arrow $E_A$ is the field produced by a negative carrier in region $A$. The blue field $E_A$ is stronger than the red field $E_A$ because the direction from $A$ to $q$ is closer to being perpendicular to the direction of the velocity of the negative charge carrier in $A$ than to that of the positive charge carrier. Similarly, the red $E_B$ and blue $E_B$ are the fields at $q$ due to the positive and negative carriers in $B$, respectively. Here, red $E_B$ is strong than blue $E_B$.

When you add the four field vectors at $q$ you get a net electric field that is upward, parallel to the current.

[EDIT: This qualitative analysis can be found in Purcell's textbook: chapter 6 in the first edition and chapter 5 in the second and third editions.]

 

[Glare]

Amazing explanation.
Thank you very much!

 

[Glare]

Here's an attempt at a qualitative explanation.

Suppose we have a very long, straight wire with current. Imagine the current is due to motion of both positive and negative charge carriers.

View attachment 346261
The positive charges are shown in red and move upward. The negative charges are blue and move downward at the same speed. The charge density is the same for the positive and negative charge. The net current is upward. The magnetic field in the region to the right of the wire points into the page. So, a positive point charge that moves away from the wire toward the right will experience a magnetic force upward:

View attachment 346262
If we go to the rest frame of the moving charge $q$, the upward force on $q$ must be due to an electric field. We want to see how this happens. In the rest frame of $q$, the wire moves toward the left. So, the positive charges in the wire move upward and to the left. The negative charges in the wire move downward and to left. This is shown in the figure below. In the small region of the wire marked $A$, the red arrow indicates the velocity of the positive charge carriers, the blue arrow denotes the velocity of the negative carriers.
View attachment 346263

A uniformly moving point charge has a radial field that is stronger in directions perpendicular to the velocity direction. This relativistic effect is very small for non-relativistic speeds, but it's important here (I think).

View attachment 346265

The pattern of the field lines for the carriers in region $A$ are shown schematically below:
View attachment 346266

Finally, consider the diagram below which shows symmetrically placed regions $A$ and $B$ of the wire. The electric fields produced by the carriers in $A$ and $B$ at the location of the charge $q$ are shown.
View attachment 346267
The red arrow $E_A$ denotes the field at $q$ produced by a positive carrier in region $A$. The blue arrow $E_A$ is the field produced by a negative carrier in region $A$. The blue field $E_A$ is stronger than the red field $E_A$ because the direction from $A$ to $q$ is closer to being perpendicular to the direction of the velocity of the negative charge carrier in $A$ than to that of the positive charge carrier. Similarly, the red $E_B$ and blue $E_B$ are the fields at $q$ due to the positive and negative carriers in $B$, respectively. Here, red $E_B$ is strong than blue $E_B$.

When you add the four field vectors at $q$ you get a net electric field that is upward, parallel to the current.

Thank you very much
Isn't this 'pancakification' effect of the electric field should also take place in the 'classical' case where the particle moves parallel to the wire, (as In addition to the length contraction effect).
Isn't these two mechanism compete each other?

 

[Sagittarius A-Star]

Isn't this 'pancakification' effect of the electric field should also take place in the 'classical' case where the particle moves parallel to the wire, (as In addition to the length contraction effect).
Isn't these two mechanism compete each other?

No. This 'pancakification' can be described as Lorentz contraction effect. The same applies to the line density of charge in the 'classical' case. In Purcell's book, the 'classical' case is calculated and gives a correct quantitative result without an "additional" effect.