# Griffiths Example 5.8

1. Apr 6, 2008

### ehrenfest

[SOLVED] Griffiths Example 5.8

1. The problem statement, all variables and given/known data
This question refers to Griffiths E and M book.

In this Example, Griffiths says "the z-component of the B-field cannot possibly depend on the direction of the current in the xy plane. (Think about it!)"

I thought about it and thought about it and thought about it and I don't see why!!?!

2. Relevant equations

3. The attempt at a solution

2. Apr 6, 2008

### Poop-Loops

Well he explains it later by saying that if you reversed the current, the B-field would point inward, which would be ridiculous.

EDIT: Never mind, I must have been thinking of a different example. No, he doesn't explain it. But the reason the Z-component of the B-field cannot depend on the direction of current is because reversing the current would give you ridiculous answers.

For example, if $$B_z$$ was uniform in the z-direction (up), reversing the current would make the B field point into the surface, which doesn't make sense.

If $$B_z$$ went like 1/r, then reversing the current would make it go like -1/r, which also doesn't make sense.

Last edited: Apr 6, 2008
3. Apr 6, 2008

### ehrenfest

Why not?!?! It seems reasonable to me that if you reverse the current, the sign of B could change!

4. Apr 6, 2008

### pam

It is an infinite sheet of current. For any point, there will be as much current on one as the other. B_z due to current on one side would be up, but on the other down.

5. Apr 6, 2008

### Poop-Loops

Yes, it does, but the point is that there is no magnetic field in the z-direction, because reversing the current would change the B-field direction.

Okay, think of it this way. It DOES have a component in the y-direction (current is in the x-direction, remember), so above the plate it is going +y, below the plate -y. Switch the current to go the opposite way, and above the plate it's -y and above it's +y. Everything still the same. If you look at it from a different angle, you can get back what you started with, right?

But if you have B going in the z-direction, then switching the current produces something completely different, not just a mirror, which doesn't make any sense at all because you're just switching the direction of current, which is totally arbitrary.

6. Apr 6, 2008

### ehrenfest

I think I see. If goes in the (+z)-direction above the plane, then it must go in the (-z)-direction below the plane by symmetry i.e. it would always point away from the plane. Switching the current would then make it point towards the plane both above and below the plane, which really doesn't make sense.