# Electromagnetics help

1. Feb 22, 2005

### robert25pl

Finding E and B fields from forces experinced by a test charge with three diferent velocities

F1=qEo(ax-ay+az) for v1=Uo*ax
F2=qEo(ax-ay-az) for v2=Uo*ay
F3=0 for v3=Uo*az

I used Lorenz force equation and by reduce E by subtracting equation 1 from 2 and eq. 3 from 2 I got

Uo(ay-ax) X B = Eo(-2az)
Uo(ay-az) X B = qEo(ax-ay-az)

So

B = C(-2az) X (ax-ay-az) = -C

Then I solved For C and for B. I not sure that I got right to this point. Can someone help me with that problem. Thanks

2. Feb 22, 2005

### Gamma

I don't understand what you have done. What is the reason for doing F1-F2 and so on.

I would have used the equation for lorenz force to 3 individual cases.

F = q( E + v x B)
qE0 ( a, -a, a) = q(Ex, Ey, Ez) + (a U0, 0, 0) X (Bx, By, Bz) ----------(1)

Write similar equations for case 2 and 3.

You have 9 equations to solve for 6 unknowns.

3. Feb 24, 2005

### robert25pl

Sorry, This is a problem.:
The forces experienced by a test charge q at a point in a region of electric and magnetic fields, respectivlely, are given with three diferent velocities, where Uo and Eo are constants. Find E and B at that point. Use Lorenz force equation.
given:
F1=qEo(ax-ay+az) for v1=Uo*ax
F2=qEo(ax-ay-az) for v2=Uo*ay
F3=0 for v3=Uo*az

From Lorenz force equation I have
qE+q*Uo(ax) X B = qEo(ax-ay+az)
qE+q*Uo(ay) X B = qEo(ax-ay-az)
qE+q*Uo(az) X B = 0

I reduce q and E by subtracting equation 1 from 2 and eq. 3 from 1. I got

Uo(ay-ax) X B = Eo(-2az)
Uo(ax-az) X B = Eo(ax-ay+az)

so if (-2az) and (ax-ay+az) is perpendicular to B then (ay-ax) and (ax-az) is parallel to B so wwe will have

B = C(-2az) X (ax-ay+az) = -C (is this right?)

where C isa proportionality constant. I solved For C and for B. I not sure that I got right to this point. I never did your way so I'm trying the way I know it can be solve. I hope You will understand now better. Thanks

4. Feb 25, 2005

### Gamma

This is wrong. If this is the case then (ay-ax) X B = 0 and (ax-az) X B = 0 which are not true.

Any way your idea is right.

This is wrong. Check again. Right hand side should have x and y. This is a cross product.

When you do this right, subtitute it in your lorentz force equatins. Equating the coefficient of the x, y, z components should give you B, and E.

5. Feb 25, 2005

### robert25pl

So is this should look like this?

B = C(-2az) X (ax-ay+az) = -C(ax-ay)

Substituding

Uo(ay-ax) X -C(ax-ay) = Eo(-2az)

-UoC(2az) = Eo(-2az)

C = Eo/Uo

Then B = (Eo/Uo)(ax-ay) Is this ok? I really need help with that? Thanks

6. Feb 25, 2005

### Gamma

B = C(-2az) X (ax-ay+az) = -2 a^2 C(x+y). Since 'a' is a constnat, say

B = D (x+y)

Substituting

Uo(ay-ax) X D(x+y) = Eo(-2az)
Uo a D (-2z) = Eo(-2az)
D = Eo/Uo

Therefore B = Eo/Uo (x+y)

Do similar thing to find E. Don't make mistake when you take the cross product.

I am getting,

E = (aEo, -aEo, 0) or in your notation E = (aEox - aEoy)

Last edited: Feb 25, 2005
7. Feb 26, 2005

### robert25pl

Thanks for help. I was reading some book and they show another way to do it. From these two equations

Uo(ay-ax) X B = Eo(-2az)
Uo(ax-az) X B = Eo(ax-ay+az)

we can use componets and solve

ax ay az
Uo -1 1 0 = Eo(-2az)
Bx By Bz

Then

Uo[Bz*ax+By*az - Bx*az+Bz*ay] = Eo(-2az)
So:
By-Bx = -Eo/Uo(az) and Bz=0

Then I will do same thing for second equation. Is this right way to do it too? Is there simplers way to do that kind problems? Thanks