Griffiths 6.5: Find Magnetic Field in Uniform Current Slab

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Homework Help Overview

The problem involves finding the magnetic field inside a uniform current slab defined by a current density \(\mathbf{J} = J_0 \hat{\mathbf{z}}\) that extends from \(x = -a\) to \(x = +a\). This is related to Griffiths Problem 6.5 in electromagnetism.

Discussion Character

  • Exploratory, Assumption checking, Conceptual clarification

Approaches and Questions Raised

  • Participants discuss the application of the right-hand corkscrew rule to determine the direction of the magnetic field, questioning its relevance for a slab versus a wire. There are attempts to relate current density to electric fields and considerations of symmetry in the magnetic field's direction.

Discussion Status

The discussion is ongoing, with participants exploring different interpretations of the magnetic field's behavior in relation to the slab's geometry. Some guidance has been provided regarding the use of the right-hand rule, but questions remain about the implications of symmetry and cancellation effects on the magnetic field direction.

Contextual Notes

Participants are navigating assumptions about the applicability of Ohm's law and the nature of the magnetic field in relation to established examples from the textbook.

ehrenfest
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[SOLVED] griffiths 6.5

Homework Statement


A uniform current density [itex]\mathbf{J} = J_0 \hat{\mathbf{z}}[/itex] fills a slab straddling the yz plane x=-a to x=+a. Find the magnetic field inside the slab.
This is part of Griffiths Problem 6.5.

Homework Equations



[tex]\nabla \times \mathbf{B} = \mu_0 \mathbf{J}[/tex]
[tex]\int \mathbf{B} \cdot d\mathbf{l} = \mu_0 I_{enc}[/tex]

The Attempt at a Solution


I first need to find the direction of B. What can I use to do that? The right-hand corkscrew rule does not seem to help here!
 
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ehrenfest said:

The Attempt at a Solution


I first need to find the direction of B. What can I use to do that? The right-hand corkscrew rule does not seem to help here!

Sure it does. Why do you think it doesn't?
 
The right-hand corkscrew rule gives you the B-field around a wire. How do you use it to get the B-field of a slab? Why do you think it does help?
 
I would approach this question by first noting that,

[tex]\underline{J}=\sigma\underline{E}[/tex]

And then using the differential form of Faraday's law,

[tex]\text{curl}\left(\underline{E}\right) = -\frac{\partial\underline{B}}{\partial t}[/tex]
 
Hootenanny said:
I would approach this question by first noting that,

[tex]\underline{J}=\sigma\underline{E}[/tex]

Wait, but that is not always true. What makes you think Ohm's law applies here?
 
ehrenfest said:
Wait, but that is not always true. What makes you think Ohm's law applies here?
Sorry, I was looking at the wrong question.

The direction of the magnetic field can definitely be found using the RH 'corkscrew' rule, as you would for a wire. Take your right-hand and place your thumb in the direction of the current density, your curling fingers indicate the direction of the magnetic field.
 
Hootenanny said:
The direction of the magnetic field can definitely be found using the RH 'corkscrew' rule, as you would for a wire. Take your right-hand and place your thumb in the direction of the current density, your curling fingers indicate the direction of the magnetic field.

OK. But in the case of the wire the B-field is circumferential. Why is it not circumferential here? Obviously there will be some cancellation here...but how do we know that after all the cancellation is done the B-field will only be in the y-direction?
 
Never mind. In Griffith's Example 5.8, he proved that the magnetic field of an infinite uniform surface current [itex]K\hat{\mathbf{x}}[/itex] flowing over the xy plane produces a magnetic field that is only in the y-direction. This is just a stack of those! We can replace x by z here due to symmetry I think.
 
Last edited:

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