##### Simplistic case: infinite wire
ems.ms.dcB.iw

Let us start with the simple case of the field \({\bf B}\) coming from an infinite wire, as given by Bwire1, rewritten here in cylindrical coordinates along the axis of the wire:

- Gr (5.39)

\[ {\bf B}_\mbox{wire} = \frac{\mu_0 I}{2\pi r} \hat{\boldsymbol \varphi} \tag{Bwire}\label{Bwire} \]

The first observation is that since the field is purely circumferential, its divergence cyl_div manifestly vanishes, \[ \nabla \cdot {\bf B}_\mbox{wire} = 0 \]

Looking for the curl, we might first want to apply cyl_curl on Bwire. The only term we should worry about is \[ \frac{1}{r} \frac{\partial}{\partial r} \left(r \times \frac{\mu_0 I}{2\pi r}\right) \] which is manifestly zero for \(r \neq 0\), but has an as-yet-unresolved \(0/0\) singular form at the origin. To treat this term correctly, (and keeping Stokes in mind), we can take a step back and calculate the line integral of \({\bf B}\) along a circular path of radius \(r\) centered on the wire. For a generic path, \(d{\bf l} = dr ~\hat{\bf r} + s d\varphi ~\hat{\boldsymbol \varphi} + dz ~\hat{\bf z}\) so for a loop encircling the wire once, \[ \oint d{\bf l} \cdot {\bf B}_\mbox{wire} = \frac{\mu_0 I}{2\pi} \int_0^{2\pi} d\varphi \frac{1}{r} r = \mu_0 I. \]

This immediately generalizes to a collection of straight wires: \[ \oint_{\cal P} d{\bf l} \cdot {\bf B} = \mu_0 I_{enc} \label{Gr(5.42)} \] where \(I_{enc}\) is the current flow through a surface \({\cal S}\) defined by the closed path \({\cal P}\):

\[ I_{enc} = \int_{\cal S} d{\bf a} \cdot {\bf J} \] Applying Stokes' theorem then immediately yields

\[ {\boldsymbol \nabla} \times {\bf B}_\mbox{wire} = \mu_0 {\bf J} \]

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Author: Jean-Sébastien Caux

Created: 2022-03-24 Thu 08:42