Tags: Rotational Symmetry / Cylindrical Coordinates / Ampere's law / Stoke's theorem / Biot-Savart's law

One of the **basic problems** of magnetostatics is the infinite wire. Using the several **symmetries** of the **current distribution** we will be able to not only find the **magnetic field** - we will also verify the "**right-hand rule**".

## Problem Statement

Calculate the **magnetic field** of an infinitely long wire with a **uniform current distribution**

in the following way:

**Explain**why the magnetic field for the given current distribution only depends on*ρ*(cylindrical coordinates), i.e. why.**B**(r) =**B**(ρ)**Additional**-**Verify the “right-hand-rule”**: - if a current distribution is generally given by*j(ρ)*(translation and rotation symmetry), show that the magnetic field can**e**_{z}**only**have an angular component*φ*.- Use the integral form of Ampère's law to calculate the
**magnetic field**everywhere.

Note that we use mostly **differential methods** to calculate the fields everywhere in this problem. In "The Magnetic Field due to a Thin Infinite Wire" we use the Biot-Savart law (integration!) to determine the magnetic field.

Nevertheless, our general solution here is also valid with respect to thin wires if we let *R → 0*.

## Hints

The **curl** of a vector field in **cylindrical coordinates** is given by

You cannot use the thin-wire Biot-Savart here. Maybe you find use for **Stoke's theorem**,

## Solution ✔

Let us determine the magnetic field for uniform current in the wire using all of the **symmetries** that are given in the system. First of all, let us find that \(\mathbf{B}\) can only depend on the (cylindrical) radial coordinate. With this result, we will easily be able to **verify the right-hand** rule for a general class of current distributions. The solution will then be a last easy step using Stoke's theorem.

## Translation and Rotation Symmetries: \(\mathbf{B}\left(\mathbf{r}\right)=\mathbf{B}\left(\rho\right)\)

**Ampere's law** relates the magnetic induction \(\mathbf{B}\left(\mathbf{r}\right)\) and the current density \(\mathbf{j}\left(\mathbf{r}\right)\): \[\nabla\times\mathbf{B}\left(\mathbf{r}\right) = \mu_{0}\mathbf{j}\left(\mathbf{r}\right)\ .\]Note that for non-magnetic media, the magnetic field \(\mathbf{H}\) is related to the magnetic induction by \(\mathbf{B}\left(\mathbf{r}\right)=\mu_{0}\mathbf{H}\left(\mathbf{r}\right)\). Then, we may also call \(\mathbf{B}\) magnetic field. The given current distribution (in cylindrical coordinates),\[\mathbf{j}\left(\mathbf{r}\right) = \begin{cases}j_{0} & \rho\leq R\\0 & \mathrm{else}\end{cases}\mathbf{e}_{z}\ ,\]obeys a **translational symmetry** - there is no change of the current along the z coordinate.

This means, that the magnetic field cannot depend on this coordinate - the current is its source! Furthermore, we find that \(\mathbf{j}\) is **rotationally symmetric** - there is no explicit dependence on the angular coordinate \(\varphi\) (remember that i.e. \(x=\rho\cos\varphi\)). Hence, also the magnetic field cannot depend on \(\varphi\)! So we know that the magnetic field has to be given by\[\mathbf{B}\left(\mathbf{r}\right) \equiv \mathbf{B}\left( \rho \right)\ .\]We may now go a little further **exploiting the symmetries** to verify the “right-hand rule”.

## Verification of the Right-Hand Rule

Let us convince ourselves that the **magnetic field can only have an angular component** if\[\mathbf{j}\left(\mathbf{r}\right) = j\left(\rho\right)\mathbf{e}_{z}\ .\]We need to consider the curl of \(\mathbf{B}\) in Ampère's law directly. In suitable cylindrical coordinates we find\[\begin{eqnarray*} \nabla\times\mathbf{B}\left(\rho\right)&=&\left(\frac{1}{\rho}{\color{red}{\partial_{\varphi}B_{z}\left(\rho\right)}}-\color{red}{\partial_{z}B_{\varphi}\left(\rho\right)}\right)\mathbf{e}_{\rho}+\left({\color{red}{\partial_{z}B_{\rho}\left(\rho\right)}}-\partial_{\rho}B_{z}\left(\rho\right)\right)\mathbf{e}_{\varphi}\\&&+\frac{1}{\rho}\left(\partial_{\rho}\left[\rho B_{\varphi}\left(\rho\right)\right]-{\color{red}{\partial_{\varphi}B_{\rho}\left(\rho\right)}}\right)\mathbf{e}_{z}\\&=&-\partial_{\rho}B_{z}\left(\rho\right)\mathbf{e}_{\varphi}+\frac{1}{\rho}\partial_{\rho}\left[\rho B_{\varphi}\left(\rho\right)\right]\mathbf{e}_{z}\ . \end{eqnarray*}\]Here we have **highlighted** terms in red that **vanish** since \(\mathbf{B}\left(\mathbf{r}\right)=\mathbf{B}\left(\rho\right)\) also holds for the slightly more general current distribution (same symmetries!).

Furthermore, our **current density** has only a component in z direction. By **Ampère's law**, the same has to hold for the curl of the magnetic field! From the latter equation, we see that the \(\mathbf{e}_{\varphi}\) component of the curl of the magnetic field has to vanish. Hence, \(\partial_{\rho}B_{z}\left(\rho\right)=0\). Also, since the magnetic field should vanish for \(\rho\rightarrow\infty\), \[B_{z}\left(\rho\right) \equiv 0\ .\]Now we can be convinced that the magnetic field can only have an angular component \(B_{\varphi}\left(\rho\right)\) for charge distributions with translational and rotational symmetry! In general, symmetries are very powerful and can be used to simplify problems to a great extend, see also "A Cylinder Shell Current and its Magnetic Field".

So, we have really **verified the right-hand rule**: a current represented by the thumb results in a magnetic field following the fingers of our right hand.

It's time to determine the magnetic field!

## Calculation of the Magnetic Field

If we integrate Ampère's law over a two-dimensional surface \(\Omega\), we can apply **Stoke's theorem** to find the integral version of Ampère's law:\[\begin{eqnarray*} \int_{\Omega}\nabla\times\mathbf{B}\left(\mathbf{r}\right)d\mathbf{A}&=&\int_{\Omega}\mu_{0}\mathbf{j}\left(\mathbf{r}\right)d\mathbf{A}\ ,\ \text{so}\\\oint_{\partial\Omega}\mathbf{B}\left(\mathbf{r}\right)d\mathbf{r}&=&\int_{\Omega}\mu_{0}\mathbf{j}\left(\mathbf{r}\right)d\mathbf{A}\ . \end{eqnarray*}\]

Of course we are clever and take a **circular surface**. Here, the boundary is given by some fixed radius \(\rho\) where the magnetic field is constant. Then,\[\begin{eqnarray*}\oint_{\partial\Omega}\mathbf{B}\left(\mathbf{r}\right)d\mathbf{r}&=&\oint_{\partial\Omega}B_{\varphi}\left(\rho\right)\rho d\varphi=2\pi\rho B_{\varphi}\left(\rho\right)\ \text{and}\\\int_{\Omega}\mu_{0}\mathbf{j}\left(\mathbf{r}\right)d\mathbf{A}&=&\mu_{0}\begin{cases}\pi\rho^{2}j_{0} & \rho\leq R\\\pi R^{2}j_{0} & \text{else}\end{cases}\ .\end{eqnarray*}\]The result for the integration over the whole wire is also the total electric current\[I = \int_{\text{wire}}\mu_{0}\mathbf{j}\left(\mathbf{r}\right)d\mathbf{A}=\pi R^{2}j_{0}\ .\]We conclude that the magnetic field is finally given as\[\begin{eqnarray*} \mathbf{B}\left(\mathbf{r}\right)&=&\mathbf{e}_{\varphi}\frac{\mu_{0}}{2\pi\rho}\begin{cases}\pi\rho^{2}j_{0} & \rho\leq R\\\pi R^{2}j_{0} & \rho>R\end{cases}\\&=&\mathbf{e}_{\varphi}\frac{\mu_{0}I}{2\pi\rho}\begin{cases}\frac{\rho^{2}}{R^{2}} & \rho\leq R\\1 & \text{else}\end{cases}\ . \end{eqnarray*}\]Note that the result of an **infinitely thin wire** is also reproduced for \(\rho\geq R\).

We can also see what we had to do if the **current distribution** was a **more complicated** function of \(\rho\). We would **have to simply integrate inside the wire to find the magnetic field there. Outside of the wire it becomes evident that the only** important parameter is indeed the **current** - the field will always look like if it was originated from an infinitely thin wire. Of course this sounds familiar if we remember "The Electric Field of a Point Charge", where the field outside of a radially symmetric charge distribution is entirely determined by the net charge.