# Bound Currents: Why Bother?

1. Sep 29, 2013

### Astrum

In Griffith's EM text, he devouts 2 pages to deriving the equation for bound currents, and for the next 4 problems, he (the solution manual) doesn't even use the equations just introduced. I question the wisdom of deriving an equation that is harder to work with than we already had.

$$\vec{A}(\vec{r}) = \frac{\mu _0}{4 \pi} ( \int _V \frac{\vec{J}_b}{r'}dV' + \oint _S \frac{\vec{K}_b}{r'}da')$$

When using this, I always get a really really ugly integral that would be too messy to work with. I guess it has some worth from a theoretical stand point, but I don't really understand why the questions that proceed this don't use the material from the same section.

2. Sep 29, 2013

### WannabeNewton

Those problems are for extremely simple systems wherein the magnetic field can be easily deduced once $\nabla \times M$ and $M \times \hat{n}$ are calculated. Given a very complicated magnetized system (indeed an arbitrarily complicated magnetized system) one cannot easily make identifications of the magnetic field in the same manner. In such a case the formula $A(r) = \frac{\mu_0}{4\pi}\int _{\mathcal{V}}\frac{\nabla \times M(r')}{|r - r'|}d\tau' + \frac{\mu_0}{4\pi}\oint _{\mathcal{S}}\frac{M\times \hat{n}}{|r - r'|}da'$ is needed for the calculation. If we are indeed dealing with a very simple magnetized system then why use a complicated formula when much simpler methods are available?

I mean you could make the same complaint about Gauss's law in integral form $\oint _{\mathcal{S}}E\cdot da = \frac{q}{\epsilon_0}$. Most problems in the textbook involve extremely simple systems with simple symmetries that allow you to never even have to perform an integral in any real sense. But most systems don't have such symmetries and you would actually have to perform the integral-just because it isn't in a textbook problem doesn't mean the formula is unneeded.

3. Sep 30, 2013

### Astrum

Yes, point taken. He just picked a strange set or problems following this section.

After thinking about it, I can see that using the equation above is easier than dealing with the "parent" equation (the equation we derived it from). $\nabla \times \vec{M}, \quad \vec{M} \times \hat{n}$ are easier to deal with than what we had before.

4. Sep 30, 2013

### WannabeNewton

Yes and much more importantly, the derivations of the quantities $j_b = \nabla \times M$ and $k_b = M \times \hat{n}$ allow us to physically interpret the bound currents. Griffiths does this in section 6.2.2.