# Griffiths 6.5

[SOLVED] griffiths 6.5

## Homework Statement

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.

## Homework Equations

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

## 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!

Ben Niehoff
Gold Member

## 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?

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

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

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

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

I would approach this question by first noting that,

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

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

Hootenanny
Staff Emeritus
Gold Member
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.

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 $K\hat{\mathbf{x}}$ 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: