# Electric field required to keep electron at rest

1. Aug 14, 2013

### boiteporte

1. The problem statement, all variables and given/known data
An electron is travelling through a magnetic field B=(Bx, By, Bz). Find an expressions for the electric field E required to keep the electron's velocity constant at v=(0,vy,0). Hence, show that E and B are perpendicular.

2. Relevant equations
F=q(v cross B)
F=qE

3. The attempt at a solution
Ok, I totally messed up this question. That's for sure. First, I decided to set the two forces equal to each other using q=-e since it is an electron. So, -e(v cross B)=eE (There are no minus signs on the right hand side, since it is going in the opposite direction).

So, E= -(v cross B). Then I had absolutely no idea what to do so the rest is probably wrong but, that's what I did (I decided to split it up in the 3 components) :

For the x-component, I got Ex=-(-vy*Bz)=vy*Bz
For the y-component, I got Ey=0
For the z-component, I got Ez=-(vyBx)=-vyBx

So, putting them altogether gives E=(vyBz,0,-vyBx). I don't think this answer is right though...

For the second part of the question, I said that the force is perpendicular to the magnetic field (as shown by the Lorentz Force equation) and since the force is parallel to the electric field, then the electric field is perpendicular to the magnetic field. I know there is a way to show it using Maxwell's Equation but the way the question is written implies that I have to use my previous answer. But I'd like some help please...

Last edited: Aug 14, 2013
2. Aug 14, 2013

### voko

There is no mistake in your derivation.

For the second part, how could you check that two vectors are perpendicular?

3. Aug 14, 2013

### boiteporte

Two vectors are perpendicular if their dot products equals 0. But, I still don't see where this is going...

4. Aug 14, 2013

### boiteporte

Sorry for my previous reply. I understand now. I have E=(vyBz,0,-vyBx) and B=(Bx,By,Bz). So, their dot products is vyBzBx+0-vyBxBz=0. So, they are perpendicular. Is this correct now?

5. Aug 14, 2013

### voko

Yes, it is all correct. Well done!