# Paricles in crossed electric and magnetic field

1. Dec 9, 2014

### iAlexN

An electron and a positron are moving in the +x-direction with the same velocity in a crossed electric and magnetic field (the fields are perpendicular). The question states it's impossible to separate them using this configuration.

The electric field is pointing in the -y-direction and the magnetic field, out of the page (+z-direction). Taking the +y-direction as positive for the electric field and $vB$, I get this:

$F_e = -e(E-vB) = e(-E+vB)$ (force on electron)
$F_p = e(-E-vB)$ (force on positron)

The cross product between the velocity (in the +x-direction) and the magnetic field (+z-direction) points the same for the electron and the positron (I think), -y-direction.

As you can see $F_e$ and $F_p$ are not equal, so the particles would be separated, which is not supposed to be possible.

But I just cannot find where I go wrong.

Thank you!

2. Dec 9, 2014

### Simon Bridge

Why is there a minus sign for the positron E field magnitude and not for the electron E field?
How did you account for the attractive force between the positron and the electron?

3. Dec 10, 2014

### Staff: Mentor

Why do you have -E in the force on the positron?

For a given E field, the forces on an electron and a positron should be in opposite directions.

4. Dec 10, 2014

### iAlexN

I agree, the E-field should have the same sign, in this situation I defined the minus y-direction as negative. I get:

$Fe=−e(-E−vB) = e(E+vB)$
$Fp=e(-E−vB)$

Supposedly (according to the question) you need not consider the attraction between the electron and the positron, just the Lorentz's Force to show this.

However, this gives that the magnetic force on the electron and positron are in the same direction as the electric force, which means they would separate? But they are not supposed to.

5. Dec 10, 2014

### Simon Bridge

Oh now I get you!
Lets see - electric field is $\vec E = -E\hat\jmath$, the magnetic field is $\vec B = B\hat k$, and the velocity is $\vec v = v\hat\imath$

Check my reasoning...
For an arbitrary charge q:
$\vec F = q\big[-E\hat\jmath + vB(\hat\imath \times \hat k = -\hat j)\big] = q(-E - vB )\hat\jmath$

For the particle to maintain a straight line path, $E+ vB=0$ ... if you fix the electric field and adjust the magnet, you see that it has to be $B=-E/v < 0$
i.e. the magnetic field needs to point the other way; well done.

Is this in the context of a CRT experiment?