# Electron force problem

If an electron is pushed up by a magnetic force, the magnetic force acts in the z direction and the electron moves in the x direction, the force is therefore in the y direction. Now if I add an electric force so that the electron carries on in the x direction. Will that electric force need to act in the negative y direction with the same magnitude as the magnetic force?

Thanks.

Related Introductory Physics Homework Help News on Phys.org
SammyS
Staff Emeritus
Homework Helper
Gold Member

If an electron is pushed up by a magnetic force, the magnetic force acts in the z direction and the electron moves in the x direction, the force is therefore in the y direction. Now if I add an electric force so that the electron carries on in the x direction. Will that electric force need to act in the negative y direction with the same magnitude as the magnetic force?

Thanks.
That's not quite correct: Let's rewrite it.

"If an electron is acted upon by a magnetic [STRIKE]force[/STRIKE] field, the magnetic [STRIKE]force[/STRIKE] field is in the z direction and the electron moves in the x direction, the force is therefore in the y direction. "

Now if I add an electric force so that the electron carries on in the x direction. Will that electric force need to act in the negative y direction with the same magnitude as the magnetic force?
The answer to the question is YES.

OK thanks. How can I use the Lorentz force law to prove that?

I have done this:

$$E = F - (V x B )/ q$$

Not sure how that equation can tell be the direction. Any help?

gneill
Mentor

The Lorentz force:

$$\vec{F} = q\left[ \vec{E} + \left( \vec{v} \times \vec{B}\right) \right]$$

Everything in the equation is a vector except for the charge, q (which does have a sign though). The "x" is the cross product. The directions of things are determined by the rules of vector manipulation.

In your case you want the net force to be zero, so you can rearrange to solve for E, as you've done. Now, when each of the vectors involved have only a single non-zero component, the expansion of the vector expression into separate component expressions becomes relatively easy (if you want to solve the problem "mechanically"). Otherwise, a little intuition about the directions that the fields must go in order to provide the required force directions on the moving, charged particle will suffice.