Understanding the Directional Independence of B-Field in Griffiths Example 5.8

In summary: 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.In summary, Griffiths says that the z-component of the B-field cannot depend on the direction of the current in the xy plane.
  • #1
ehrenfest
2,020
1
[SOLVED] Griffiths Example 5.8

Homework Statement


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!?

Homework Equations


The Attempt at a Solution

 
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  • #2
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 [tex]B_z[/tex] 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 [tex]B_z[/tex] went like 1/r, then reversing the current would make it go like -1/r, which also doesn't make sense.
 
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  • #3
Poop-Loops said:
For example, if [tex]B_z[/tex] 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 [tex]B_z[/tex] went like 1/r, then reversing the current would make it go like -1/r, which also doesn't make sense.

Why not?? It seems reasonable to me that if you reverse the current, the sign of B could change!
 
  • #4
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
ehrenfest said:
Why not?? It seems reasonable to me that if you reverse the current, the sign of B could change!

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
Poop-Loops said:
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.

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.
 

1. What is Griffiths Example 5.8?

Griffiths Example 5.8 is a problem presented in the textbook "Introduction to Electrodynamics" by David J. Griffiths. It involves finding the electric field inside and outside of a charged spherical shell using Gauss's law.

2. What is the purpose of Griffiths Example 5.8?

The purpose of Griffiths Example 5.8 is to demonstrate the application of Gauss's law to a specific problem, and to show how the electric field behaves inside and outside of a spherical shell of charge.

3. What are the key equations used in Griffiths Example 5.8?

The key equations used in Griffiths Example 5.8 are Gauss's law, Coulomb's law, and the definition of electric field. These equations are used to calculate the electric field at different points inside and outside of the charged spherical shell.

4. What are the main steps to solving Griffiths Example 5.8?

The main steps to solving Griffiths Example 5.8 are: 1) drawing a diagram of the problem, 2) determining the charge distribution and symmetries present, 3) applying Gauss's law to find the electric field inside and outside of the spherical shell, and 4) using Coulomb's law to calculate the electric field at the center of the shell.

5. What are the applications of Griffiths Example 5.8 in real life?

Griffiths Example 5.8 has applications in understanding the behavior of electric fields in spherical geometries, which can be useful in many real-life situations such as analyzing the electric field around a charged particle or a spherical capacitor. It also highlights the importance of symmetries in simplifying electrostatic problems.

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