- #1

jack476

- 328

- 125

- Homework Statement
- For a distribution of volume currents ##\vec{J}^\prime(\vec{r{^\prime}})## distributed throughout a finite volume, show that the Biot-Savart law can be written as ##\vec{B}(\vec{r}) = \frac{\mu_0}{4\pi}\int_{V^{\prime}}\frac{\nabla^\prime \times \vec{J}(\vec{r}^\prime)}{R}d\tau^\prime##. The source currents are steady and divergenceless.

- Relevant Equations
- The vector ##\vec{r}## is a field point, the vector ##\vec{r}^\prime## is a source point, and ##\vec{R} = \vec{r}-\vec{r}^\prime##.

When ##f(\vec{R})## and ##\vec{F}(\vec{R})## are functions that depend only on ##\vec{R}## the following identities hold: ##\nabla f(\vec{R}) = -\nabla^\prime f(\vec{R}), \nabla\cdot\vec{F}(\vec{R}) = -\nabla^\prime\cdot\vec{F}(\vec{R}), \nabla \times \vec{F}(\vec{R}) = -\nabla^\prime \vec{F}(\vec{R})##

Since ##\vec{R}/R^3 = -\nabla(\frac{1}{R}) = \nabla^\prime(\frac{1}{R})##, the standard form of the Biot-Savart law for volume currents can be re-written as: $$\frac{\mu_0}{4\pi}\int\limits_{V^\prime}\frac{\vec{J}^\prime (\vec{r}^\prime)\times\vec{R}}{R^3}d\tau^\prime = \frac{\mu_0}{4\pi}\int\limits_{V^\prime}\vec{J}^\prime (\vec{r}^\prime)\times\nabla^\prime\left(\frac{1}{R}\right)d\tau^\prime$$

By using the vector identity ##\nabla \times (f\vec{F}) = (\nabla\times \vec{F})f -\vec{F}\times(\nabla f)## the RHS can be re-written, resulting in:

$$\frac{\mu_0}{4\pi}\int\limits_{V^\prime}\frac{\vec{J}^\prime \times\vec{R}}{R^3}d\tau^\prime = \frac{\mu_0}{4\pi}\int\limits_{V^\prime}\frac{\nabla^\prime \times \vec{J}^\prime}{R}d\tau^\prime - \frac{\mu_0}{4\pi}\int\limits_{V^\prime}\nabla^\prime\times\left(\frac{J^\prime}{R}\right)d\tau^\prime$$

So the problem is solved if:

$$ \int\limits_{V^\prime}\nabla^\prime\times\left(\frac{\vec{J}^\prime}{R}\right)d\tau^\prime = 0$$

And I am at a complete loss as to how to prove that. Is there some reason for why ##\vec{J}/R## should be a conservative field even though ##\vec{J}## isn't?

By using the vector identity ##\nabla \times (f\vec{F}) = (\nabla\times \vec{F})f -\vec{F}\times(\nabla f)## the RHS can be re-written, resulting in:

$$\frac{\mu_0}{4\pi}\int\limits_{V^\prime}\frac{\vec{J}^\prime \times\vec{R}}{R^3}d\tau^\prime = \frac{\mu_0}{4\pi}\int\limits_{V^\prime}\frac{\nabla^\prime \times \vec{J}^\prime}{R}d\tau^\prime - \frac{\mu_0}{4\pi}\int\limits_{V^\prime}\nabla^\prime\times\left(\frac{J^\prime}{R}\right)d\tau^\prime$$

So the problem is solved if:

$$ \int\limits_{V^\prime}\nabla^\prime\times\left(\frac{\vec{J}^\prime}{R}\right)d\tau^\prime = 0$$

And I am at a complete loss as to how to prove that. Is there some reason for why ##\vec{J}/R## should be a conservative field even though ##\vec{J}## isn't?