Pre-Quantum Electrodynamics

Infinite straight wire: calculate line integral of \({\bf B}\) along circular path of radius \(s\) centered on wire: from \ref{Gr(5.36)}, \[ \oint {\bf B} \cdot d{\bf l} = \oint \frac{\mu_0 I}{2\pi s} dl = \mu_0 I. \] Any loop will do: \[ {\bf B} = \frac{\mu_0 I}{2\pi s} \hat{\boldsymbol \varphi} \label{Gr(5.41)} \] and \(d{\bf l} = ds \hat{\bf s} + s d\varphi \hat{\boldsymbol \varphi} + dz \hat{\bf z}\) so for a loop encircling the wire once, \[ \oint {\bf B} \cdot d{\bf l} = \frac{\mu_0 I}{2\pi} \oint \frac{1}{s} s d\varphi = \mu_0 I. \]

For a general collection of straight wires: \[ \oint_{\cal P} {\bf B} \cdot d{\bf l} = \mu_0 I_{enc} \label{Gr(5.42)} \] where \(I_{enc}\) is the current flow through a surface \({\cal S}\) defined by the closed path \({\cal P}\): \[ I_{enc} = \int_{\cal S} {\bf J} \cdot d{\bf a} \label{Gr(5.43)} \] Applying Stokes' theorem: