Magnetic Field of Uniformly Magnetized Infinite Slab

[sammydafish]

Homework Statement

Find the magnetic field above a uniformly magnetized infinite slab of thickness 2d using the Ampere's Law and the magnetic vector potential.

Relevant Equations

$$\oint \mathbf{B} \cdot d\mathbf{l} = \mu_0 I_{\text{enc}}$$

$$\mathbf{A}(\mathbf{r}) = \frac{\mu_0}{4\pi} \int \frac{\mathbf{M}(\mathbf{r}') \times (\mathbf{r} - \mathbf{r}')}{|\mathbf{r} - \mathbf{r}'|^3} \, d\tau'$$

The bound surface current is $$\mathbf{K}_b = \mathbf{M} \times \hat{n}$$ which gives $$\mathbf{K}_b = {M} \hat{x}$$
Applying ampere's law: $\oint \mathbf{B} \cdot d\mathbf{l} = \mu_0 I_{\text{enc}}$ to the top surface, which encloses the bound surface current yields: $\mathbf{B} = \mu_0 {K}_b$
This is in alignment with Griffiths Electrodynamics 4th edition example 5.8 for a surface current.

Now,
$$\mathbf{A}(\mathbf{r}) = \frac{\mu_0}{4\pi} \int \frac{\mathbf{M}(\mathbf{r}') \times (\mathbf{r} - \mathbf{r}')}{|\mathbf{r} - \mathbf{r}'|^3} \, d\tau'$$

Since the magnetization is constant, I only need to solve:
$$\int \frac{(\mathbf{r} - \mathbf{r}')}{|\mathbf{r} - \mathbf{r}'|^3} \, d\tau'$$

This is tedious and not enlightening. Therefore, I consider a different situation: A uniformly charged infinite slab of thickness 2d (Griffiths Problem 2.17), apply Gauss' Law for the electric field and get:
$$\mathbf{E} = \frac{\rho d}{\epsilon_0} \hat{z}$$
Which of course must be equal to:
$$ \mathbf{E} = \frac{1}{4\pi\epsilon_0} \rho \int \frac{(\mathbf{r} - \mathbf{r}')}{|\mathbf{r} - \mathbf{r}'|^3} d\tau' $$

Therefore:
$$\int \frac{(\mathbf{r} - \mathbf{r}')}{|\mathbf{r} - \mathbf{r}'|^3} \, d\tau' = {4 \pi d} \hat{z}$$

Which means that:

$$\mathbf{A}(\mathbf{r}) = \frac{\mu_0}{4\pi} \mathbf{M} \times {4 \pi d} \hat{z}$$

If $\mathbf{M}$ is in the y direction:

$$\mathbf{A}(\mathbf{r}) = {\mu_0 d M} \hat{x}$$

Taking the curl of this vector potential to find B yields zero as it is constant which disagrees with Ampere's Law from above.

So I have made a mistake somewhere and cannot see it. Can anyone please provide some insight? While the solution is important, I really want to understand where I am making a mistake in this method. This method works perfectly for the uniformly magnetized/charged sphere and cylinder so I see no reason why it shouldn't work here.

 

[kuruman]

Since the magnetization is constant, I only need to solve:
$$\int \frac{(\mathbf{r} - \mathbf{r}')}{|\mathbf{r} - \mathbf{r}'|^3} \, d\tau'$$

This is tedious and not enlightening.

You need to solve what? I don't see any kind of equation that could be solved.
You need to transform the right side of the equation
$$\mathbf B(\mathbf r)=\mathbf{\nabla}\times \left[ \frac{\mu_0}{4\pi} \int \frac{\mathbf{M}(\mathbf{r}') \times (\mathbf{r} - \mathbf{r}')}{|\mathbf{r} - \mathbf{r}'|^3} \, d\tau'\right].$$ You may not find it enlightening, but it's the way it's done ussing the vector identity $$\mathbf {\nabla}\times(\mathbf F\times\mathbf G)= (\mathbf{\nabla}\cdot \mathbf G)\mathbf F-(\mathbf{\nabla}\cdot\mathbf F)\mathbf G+(\mathbf G\cdot\mathbf{\nabla})\mathbf F-(\mathbf F\cdot\mathbf{\nabla})\mathbf G.$$Of course the del operator operates on unprimed coordinates only and can be taken inside the integral.

 

[sammydafish]

The integral I mentioned is solved repeatedly in any introduction to electrodynamics course before students learn about gauss' law and potentials. See Griffiths 4th edition Chapter 2 for various configurations which require solving that integral.

I understand it can be done the way you suggest, but my question is what exactly is wrong with the methodology I presented here.

 

[kuruman]

See Griffiths 4th edition Chapter 2 for various configurations which require solving that integral.

These various configurations in chapter 2 of Griffiths use Gauss's law. As you already know, in differential form it is Maxwell's equation $$\mathbf {\nabla}\cdot \mathbf E=\frac{\rho}{\epsilon_0}.\tag 1$$It says that the source of a diverging electric field at a point in space is the proportional to the volume charge density at that point. It is used to find electric fields given a certain $\rho(\mathbf r)$, not integrals.

If you want to draw an analogy between electric and magnetic fields, you need to be careful and justify your conclusion

Which means that:$$\mathbf{A}(\mathbf{r}) = \frac{\mu_0}{4\pi} \mathbf{M} \times {4 \pi d} \hat{z}$$

As you have already found out

Taking the curl of this vector potential to find B yields zero as it is constant which disagrees with Ampere's Law from above.

You went wrong in assuming that you can just replace symbols in the electric field case with corresponding symbols in the magnetic field case because "the integral is the same in the two cases".

My point is that the result for the electric field that you carried over to a magnetic field is consistent with equation (1) in which clearly $\rho \neq 0.$ Magnetic fields are consistent with Maxwell's equation $$\mathbf {\nabla}\cdot \mathbf B= 0. \tag 2$$By carrying the electric field result over to the magnetic field case, you essentially threw out the baby with the bath water and got zero for the magnetic field.

Had you done the integral the way I suggested, you would have found the general expression for the magnetic field $$\mathbf B(\mathbf r)= -\mu_0\mathbf{\nabla}\varphi^*(\mathbf r)+\mu_0\mathbf M(\mathbf r)\tag 3$$ where $\varphi^*(\mathbf r)$ is the magnetic scalar potential. Applying equation (2) to equation (3) where $\mathbf M(\mathbf r)$ is uniform (non-zero inside and zero outside) gives $$\nabla^2\varphi^*(\mathbf r)=0.\tag 4$$ The "baby" that you threw out is the magnetic scalar potential. Now you can draw an analogy with the electric field case and proceed to solve Laplace's equation (4) as if you had an electric field but with boundary conditions appropriate to a magnetic field.

 

[sammydafish]

Ok, I see. That's interesting. So this means there is something special about the slab/geometry?

I applied my method above for the E field outside a uniformly charged sphere is given by gauss' law:

$$\mathbf{E} = \frac{\rho R^{3}}{3 \epsilon_0 r^{2}} \hat{r}$$

Where again,

$$ \mathbf{E} = \frac{1}{4\pi\epsilon_0} \rho \int \frac{(\mathbf{r} - \mathbf{r}')}{|\mathbf{r} - \mathbf{r}'|^3} d\tau' $$

Therefore,


$$\frac{R^{3}}{3 \epsilon_0 r^{2}} \hat{r} = \frac{1}{4\pi\epsilon_0}\int \frac{(\mathbf{r} - \mathbf{r}')}{|\mathbf{r} - \mathbf{r}'|^3} d\tau' $$

Now applying this to the original magnetic vector potential, we arrive at:

$$\mathbf{A}(\mathbf{r}) = {\epsilon_0 \mu_0} \mathbf{M} \times \frac{R^{3}}{3 \epsilon_0 r^{2}} \hat{r}$$

If M is in the z direction, we get:

$$\mathbf{A}(\mathbf{r}) = \frac{\mu_0 R^{3}}{3 r^2}{{M}\sin{\theta} \hat{\phi}
}$$

If you then take the curl of this vector potential and apply the definition of magnetization in terms of the magnetic dipole moment per unit volume you get:

$$\mathbf{B}(\mathbf{r}) = \frac{\mu_0 m}{4 \pi r^3} \left[ 2 \cos{\theta} \hat{r} + \sin{\theta} \hat{\theta} \right] $$

Which is the dipole field I expected to get. This method also works perfectly inside the sphere. As well as for the uniformly magnetized cylinder (which I can show if you like). So is it really just by coincidence that it works without invoking the magnetic scalar potential and magnetic charge densities? That seems odd but I suppose I could accept it.

 

[TSny]

$$\mathbf{A}(\mathbf{r}) = \frac{\mu_0}{4\pi} \int \frac{\mathbf{M}(\mathbf{r}') \times (\mathbf{r} - \mathbf{r}')}{|\mathbf{r} - \mathbf{r}'|^3} \, d\tau'$$

This integral automatically includes the contributions from the surface currents on the top and bottom of the slab. Since we know both surfaces produce $\mathbf B = 0$ outside the slab, it makes sense that you get an expression for $\mathbf{A}$ outside the slab whose curl is zero.

If you calculate the integral for $\mathbf r$ inside the slab, you will get an expression for $\mathbf A(\mathbf r)$ whose curl yields the correct nonzero $\mathbf B$ inside the slab.

 

[sammydafish]

This integral automatically includes the contributions from the surface currents on the top and bottom of the slab. Since we know both surfaces produce $\mathbf B = 0$ outside the slab, it makes sense that you get an expression for $\mathbf{A}$ outside the slab whose curl is zero.

If you calculate the integral for $\mathbf r$ inside the slab, you will get an expression for $\mathbf A(\mathbf r)$ whose curl yields the correct nonzero $\mathbf B$ inside the slab.

Yes, that's it. I was missing the fact that the vector potential equation includes both the top and bottom automatically while in Ampere's Law I chose to enclose just the current in the top surface. Upon enclosing both the top and bottom surface I do of course get zero B-field.