# Griffiths 6.5

1. Apr 29, 2008

### ehrenfest

[SOLVED] griffiths 6.5

1. The problem statement, all variables and given/known data
A uniform current density $\mathbf{J} = J_0 \hat{\mathbf{z}}$ 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.

2. Relevant equations

$$\nabla \times \mathbf{B} = \mu_0 \mathbf{J}$$
$$\int \mathbf{B} \cdot d\mathbf{l} = \mu_0 I_{enc}$$

3. 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!

2. Apr 29, 2008

### Ben Niehoff

Sure it does. Why do you think it doesn't?

3. Apr 29, 2008

### ehrenfest

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?

4. Apr 29, 2008

### Hootenanny

Staff Emeritus
I would approach this question by first noting that,

$$\underline{J}=\sigma\underline{E}$$

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

$$\text{curl}\left(\underline{E}\right) = -\frac{\partial\underline{B}}{\partial t}$$

5. Apr 29, 2008

### ehrenfest

Wait, but that is not always true. What makes you think Ohm's law applies here?

6. Apr 29, 2008

### Hootenanny

Staff Emeritus
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.

7. Apr 29, 2008

### ehrenfest

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?

8. Apr 29, 2008

### ehrenfest

Never mind. In Griffith's Example 5.8, he proved that the magnetic field of an infinite uniform surface current $K\hat{\mathbf{x}}$ 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: Apr 29, 2008