# Magnetic field due to straight wire on the line itself

Tags:
1. Aug 27, 2015

### Hithesh

Is magnetic field due to straight wire on the line of the wire itself zero? If yes, give mathematical proof.

2. Aug 27, 2015

### Korak Biswas

It follows from Biot-Savart law.

3. Aug 28, 2015

### vanhees71

If you have a infinitely thin wire, the magnetic field at the wire is infinite, but that's, because there is no such thing. It is also much easier to solve the stationary Maxwell equations (DC case) directly and not to use Biot-Savart's Law. Just use the ansatz
$$\vec{j}=\sigma \vec{E}=\begin{cases} \frac{I}{A} \vec{e}_z & \text{for} \quad \rho<a, \\ 0 & \text{for} \quad \rho >a. \end{cases}$$
with $I=\text{const}$ the current and $A=\pi a^2$ the cross-sectional area of the wire. I assume that the wire is an infinitely long cylinder of radius $a$ and use cylinder coordinates $(\rho,\varphi,z)$.

In the non-relativistic limit for the current flow in the wire (which is very accurate since the drift velocities of the electrons are very slow, on the order of $10^{-3} \text{m/s}$) all we have to do is to solve the Poisson equation for the vector potential, which reads (in Coulomb gauge)
$$-\Delta \vec{A}=\vec{\nabla} \times (\vec{\nabla} \times \vec{A})=\frac{1}{c} \vec{j}.$$
For symmetry reasons we can make the ansatz
$$\vec{A}=A_z(r) \vec{e}_z.$$
Then the Poisson equation gets
$$\Delta \vec{A}=\frac{1}{\rho} \frac{\mathrm{d}}{\mathrm{d} \rho} \left (\rho \frac{\mathrm{d} A_z}{\mathrm{d} \rho} \right )=-\frac{j_z}{c}.$$
This we can easily solve by successive integration. Writing a prime for the derivative wrt. $\rho$ the equation reads for $\rho<a$
$$(\rho A_z')'=-\frac{I}{\pi c a^2} \rho \; \Rightarrow \; A_z'=-\frac{I}{2 \pi c a^2} \rho+\frac{C_1}{\rho},$$
where $C_1$ is an arbitrary constant of integration. Integrating again gives
$$A_z=-\frac{I}{4 \pi c a^2} \rho^2 + C_1 \ln \left (\frac{\rho}{a} + C_2 \right).$$
Here $C_2$ is another integration constant. I have put another constant under the log to make the argument of the log dimensionless (choosing the radius of the cylinder as a scale in the log is just for convenience). For the physical quantities it's irrelevant anyway, because now we have to work in the various boundary conditions. First of all the vector potential must be non-singular everywhere, and thus we must set $C_1=0$, because the log diverges for $\rho \rightarrow 0$.

Then we get for the magnetic field
$$\vec{B} = \vec{\nabla} \times \vec{A}=-\vec{e}_{\varphi} A_z'=\frac{I}{2 \pi c}\rho \quad \text{for} \quad \rho<a.$$
For $\rho>a$ we can set simply $I=0$ in the above general solution. Then we have
$$A_z=C_1 \ln \left (\frac{\rho}{a} \right ) +C_2.$$
$$\vec{B}=-A_z' \vec{e}_{\varphi}=-\frac{C_1}{\rho}.$$
$$C_1=-\frac{I}{2 \pi c a^2}.$$
$$\vec{B}=\frac{I}{2 \pi c \rho} \vec{e}_{\varphi} \quad \text{for} \quad \rho \geq a.$$
As you see, if you make $a \rightarrow 0^+$ the field becomes infinite at the cylinder axis. For a wire of finite radius, no singularities occur (except the kink in the field at the cylinder surface which is due to the approximation to make $\vec{j}$ jumping abruptly to 0 there).