## The Magnetic Field due to a Thin Infinite Wire

We will directly use the Biot-Savart law to calculate the magnetic field of a thin wire via integration. The problem shall provide us with some intuition of Biot-Savart which we will also derive.

## Problem Statement

In The Magnetic Field of an Infinite Wire we discuss the field of a not necessarily thin wire with a constant current $$j_{0}\mathbf{e}_{z}$$ and discuss generalizations $$j_{0}\rightarrow j\left(\rho\right)$$. We made our calculations directly from the differential form of Ampère's law. If we assume a very thin wire, i.e. a current of the form$\mathbf{j}\left(\mathbf{r}\right) = I\delta\left(\rho\right)\mathbf{e}_{z}$in cylindrical coordinates, we may also calculate the magnetic field by direct integration via Biot-Savart's law,$\mathbf{B}\left(\mathbf{r}\right) = \frac{\mu_{0}I}{4\pi}\int\frac{d\mathbf{s}\times\left(\mathbf{r}-\mathbf{r}\left(s\right)\right)}{\left|\mathbf{r}-\mathbf{r}\left(s\right)\right|^{3}}\ ,$where the integral is over the curve of the thin wire.

• Calculate the magnetic field $$\mathbf{B}\left(\mathbf{r}\right)$$ for the thin wire, infinitely extended in $$z$$ using Biot-Savart law.
• Extra: Derive the Biot-Savart law from the connection of the vector potential to the current.

## Hints

Even though the magnetic field of the infinite thin wire is a standard problem, draw a schematic.

Remember that from $$\Delta\mathbf{A}\left(\mathbf{r}\right)=-\mu_{0}\mathbf{j}\left(\mathbf{r}\right)$$ the vector potential is given by $\mathbf{A}\left(\mathbf{r}\right) = \frac{\mu_{0}}{4\pi}\int\frac{\mathbf{j}\left(\mathbf{r}^{\prime}\right)}{\left|\mathbf{r}-\mathbf{r}^{\prime}\right|}dV^{\prime}\ ,$just as in electrostatics.

## Show Solution

Since we are scientists, we shall convince ourselves first that Biot-Savart's law is indeed correct. After that we will use it to determine the field of the thin wire.

## Derivation of the Biot-Savart Law

There exist lots of different derivations of the Biot-Savart law. A lot of them use relations of differential elements $$d\mathbf{s}$$ causing differential magnetic fields $$d\mathbf{B}\left(\mathbf{r}\right)$$. Let us try here an approach inspired by a proper curve parametrization $$\mathbf{r}\left(s\right)$$.

We start at the well-known$\mathbf{A}\left(\mathbf{r}\right) = \frac{\mu_{0}}{4\pi}\int\frac{\mathbf{j}\left(\mathbf{r}^{\prime}\right)}{\left|\mathbf{r}-\mathbf{r}^{\prime}\right|}dV^{\prime}\ ,$providing us a relation of the vector potential to the current density. For thin wires, we can explain $$\mathbf{j}\left(\mathbf{r}\right)$$ as an integral over the positions $$\mathbf{r}\left(s\right)$$ of the wire,$\begin{eqnarray*} \mathbf{j}\left(\mathbf{r}\right)&=&I\,\int\frac{\partial\mathbf{r}\left(s\right)}{\partial s}/\left|\frac{\partial\mathbf{r}\left(s\right)}{\partial s}\right|\delta\left(\mathbf{r}-\mathbf{r}\left(s\right)\right)ds\\&=&I\,\int\mathbf{T}\left(s\right)\delta\left(\mathbf{r}-\mathbf{r}\left(s\right)\right)ds\ . \end{eqnarray*}$Here, $$\mathbf{T}\left(s\right)$$ is the normalized tangential vector. Then we find for the vector potential$\begin{eqnarray*} \mathbf{A}\left(\mathbf{r}\right)&=&\frac{\mu_{0}I}{4\pi}\int\frac{\mathbf{T}\left(s\right)\delta\left(\mathbf{r}^{\prime}-\mathbf{r}\left(s\right)\right)}{\left|\mathbf{r}-\mathbf{r}^{\prime}\right|}dV^{\prime}ds\\&=&\frac{\mu_{0}I}{4\pi}\int\frac{d\mathbf{s}}{\left|\mathbf{r}-\mathbf{r}\left(s\right)\right|}\ , \end{eqnarray*}$since we can change the order of integration and $$\mathbf{T}\left(s\right)ds=d\mathbf{s}$$ is basically the definition of a curve integration. Now we are almost ready. We just have to take the curl of the vector potential to find the magnetic field:$\begin{eqnarray*} \mathbf{B}\left(\mathbf{r}\right)=\nabla\times\mathbf{A}\left(\mathbf{r}\right)&=&\frac{\mu_{0}I}{4\pi}\int\nabla\times\frac{d\mathbf{s}}{\left|\mathbf{r}-\mathbf{r}\left(s\right)\right|}\ ,\\\nabla\times\frac{d\mathbf{s}}{\left|\mathbf{r}-\mathbf{r}\left(s\right)\right|}&=&\frac{1}{\left|\mathbf{r}-\mathbf{r}\left(s\right)\right|}\left(\nabla\times d\mathbf{s}\right)-d\mathbf{s}\times\nabla\frac{1}{\left|\mathbf{r}-\mathbf{r}\left(s\right)\right|}\\&=&+d\mathbf{s}\times\frac{\mathbf{r}-\mathbf{r}\left(s\right)}{\left|\mathbf{r}-\mathbf{r}\left(s\right)\right|^{3}}\ . \end{eqnarray*}$This leaves us in the end with the Biot-Savart law:$\mathbf{B}\left(\mathbf{r}\right) = \frac{\mu_{0}I}{4\pi}\int\frac{d\mathbf{s}\times\left(\mathbf{r}-\mathbf{r}\left(s\right)\right)}{\left|\mathbf{r}-\mathbf{r}\left(s\right)\right|^{3}}\ .$Now let us apply our finding!

## The Magnetic Field of the Thin Wire

For the thin and infinitely long wire we may very well parametrize the wire as $$\mathbf{r}\left(s\right)=s\mathbf{e}_{z}$$, $$s\in\left(-\infty,\infty\right)$$. Then the Biot-Savart integration takes the form$\mathbf{B}\left(\mathbf{r}\right) = \frac{\mu_{0}I}{4\pi}\int\frac{\mathbf{e}_{z}\times\left(\mathbf{r}-s\mathbf{e}_{z}\right)}{\left|\mathbf{r}-s\mathbf{e}_{z}\right|^{3}}ds\ .$Now, the term $$\mathbf{e}_{z}\times s\mathbf{e}_{z}$$ vanishes and we have to think what $$\mathbf{r}$$ actually means. Without any restriction we may say that $$\mathbf{r}=\rho\mathbf{e}_{\rho}$$, thus somewhere in the x-y-plane: Because of the obvious translational symmetry of the current distribution in z direction and the rotational symmetry in $$\varphi$$ direction, the field must obey the same symmetries. Then,$\begin{eqnarray*} \mathbf{B}\left(\rho\mathbf{e}_{\rho}\right)&=&\frac{\mu_{0}I}{4\pi}\int\frac{\mathbf{e}_{z}\times\rho\mathbf{e}_{\rho}}{\left|\rho\mathbf{e}_{\rho}-s\mathbf{e}_{z}\right|^{3}}ds\\&=&\frac{\mu_{0}I}{4\pi}\mathbf{e}_{\varphi}\int\frac{\rho}{\left(\rho^{2}+s^{2}\right)^{3/2}}ds\ . \end{eqnarray*}$This is a standard integral and we may solve it with the standard substitution$\begin{eqnarray*} \tan\alpha&=&s/\rho\ \text{, then}\\ds&=&\rho\frac{d\alpha}{\cos^{2}\alpha}\ \text{and}\\\mathbf{B}\left(\rho\mathbf{e}_{\rho}\right)&=&\frac{\mu_{0}I}{4\pi}\mathbf{e}_{\varphi}\int_{-\pi/2}^{\pi/2}\frac{\rho}{\rho^{3}\left(1+\tan^{2}\alpha\right)^{3/2}}\frac{\rho}{\cos^{2}\alpha}d\alpha\\&=&\frac{\mu_{0}I}{4\pi\rho}\mathbf{e}_{\varphi}\int_{-\pi/2}^{\pi/2}\cos\alpha d\alpha\ . \end{eqnarray*}$The last integral equals to 2. Then, since the solution cannot depend on the $$z$$ position as we have discussed before, we find the magnetic field of the thin and infinitely long wire to be $\mathbf{B}\left(\mathbf{r}\right) = \frac{\mu_{0}I}{2\pi\rho}\mathbf{e}_{\varphi}\ .$ This is of course exactly the result we already calculated for a not necessarily thin wire with net current $$I$$. This does not surprise us from our experience in electrostatics. There, the field of spherically symmetric charge distributions is that of a point charge outside of the charge distribution. Here, we may find a corresponding principle. Let us integrate Ampère's law over a certain surface $$A$$ including all of the current density. Then, we find$\begin{eqnarray*} \int_{A}\nabla\times\mathbf{B}\left(\mathbf{r}\right)d\mathbf{A}&=&\int_{\partial A}\mathbf{B}\left(\mathbf{r}\right)d\mathbf{r}\\&=&\int_{A}\mu_{0}\mathbf{j}\left(\mathbf{r}\right)d\mathbf{A}\equiv I \end{eqnarray*}$using Stoke's law. So the magnetic field on a closed loop only depends on the current through the area of that loop.