# Ampere's Law using the Biot Savart Law

## Main Question or Discussion Point

Is it possible to derive Ampere's Law(circuital) using the Biot Savart Law and elementary calculus?

Thanks

Related Other Physics Topics News on Phys.org
Ampere's Law is equivalent to the conservation of charge i.e. the continuity equation.

Can you provide the derivation please..

thanks

Defennder
Homework Helper
You can derive it with vector calculus, but definitely not only with elementary calculus. But with elementary calculus it's possible to derive a simple special case of Ampere's circuital law, namely that for a closed path around a current carrying wire. There may be other (highly-symmetric) configurations for which Ampere's law can be derived from Biot-Savart, but I can't think of any at present.

For the case of the current carrying wire, the magnetic field at a distance r from the wire can be found by Biot-Savart law and is given as $$\frac{I}{2\pi r}$$. This itself can be derived by considering a particular setup, that of a current flowing upwards in a wire aligned along the z-axis towards z. By the Biot-Savart law:

$$d\textbf{B} = \frac{\mu_0 Id\textbf{L} \times \textbf{r}}{4\pi r^3}$$.
For this setup, $$d\textbf{L} = dz \textbf{k} \ \mbox{and} \ \textbf{r} = x\textbf{i} + y\textbf{j} + z\textbf{k}$$ and this makes the numerator of the dB expression $$(x\textbf{j} - y\textbf{i})dz$$ by the cross product.

So integrating the above from -infinity to +infinity along the z-axis of current flow gives:

$$\frac{I\mu_0}{4\pi} \int_{-\infty}^{\infty} \frac{-y}{(x^2+y^2+z^2)^{3/2}} \textbf{i} + \int_{-\infty}^{\infty} \frac{x}{(x^2+y^2+z^2)^{3/2}} \textbf{j}$$

This reduces to the following after substituting for the limits of the integral expression:

$$\frac{I\mu_0}{2\pi} \left( -\frac{y}{x^2+y^2} \textbf{i} + \frac{x}{x^2+y^2} \textbf{j}\right)$$.

And the magnitude of the vector to be $$\frac{I\mu_0}{2\pi r}$$, where $$r=\sqrt{x^2+y^2}$$.

So, with this one can prove Ampere's circuital law for the simple case of a current carrying wire. Starting with $$\oint \textbf{B} \cdot d\textbf{r}$$, note that the value of magnetic flux density does not change with the line integral around the wire at the same radius, so that means we can take B outside of the integral and the closed path integral reduces to the circumference of a circle about the wire. So this means that $$\textbf{B} \oint d\textbf{r} = B (2\pi r)$$. And with the expression for B as derived above, $$\frac{I\mu_0}{2\pi r} 2 \pi r = I\mu_0$$.

You can find a clearer and better derivation for this in any introductory physics textbook.

clem