Pre-Quantum Electrodynamics

\[ {\bf B} ({\bf r}) = \frac{\mu_0}{4\pi} \int_{\cal V} d\tau' \frac{{\bf J}({\bf r}') \times ({\bf r} - {\bf r}')}{|{\bf r} - {\bf r}'|^3} \label{Gr(5.45)} \] Apply the divergence: \[ {\boldsymbol \nabla} \cdot {\bf B} ({\bf r}) = \frac{\mu_0}{4\pi} \int_{\cal V} d\tau' {\boldsymbol \nabla} \cdot \left(\frac{{\bf J}({\bf r}') \times ({\bf r} - {\bf r}')}{|{\bf r} - {\bf r}'|^3} \right) \label{Gr(5.46)} \] Note that we can write (using div1or) \(\frac{{\bf r} - {\bf r}'}{|{\bf r} - {\bf r}'|^3} = -{\boldsymbol \nabla} \frac{1}{|{\bf r} - {\bf r}'|}\). Substituting this in and using product rule number 6, \[ {\boldsymbol \nabla} \cdot ({\bf A} \times {\bf B}) = {\bf B} \cdot ({\boldsymbol \nabla} \times {\bf A}) - {\bf A} \cdot ({\boldsymbol \nabla} \times {\bf B}), \] we get

\begin{align} {\boldsymbol \nabla} \cdot \left(\frac{{\bf J}({\bf r}') \times ({\bf r} - {\bf r}')}{|{\bf r} - {\bf r}'|^3} \right) &= -{\boldsymbol \nabla} \cdot \left( {\bf J}({\bf r}') \times {\boldsymbol \nabla} \frac{1}{|{\bf r} - {\bf r}'|} \right) \nonumber \\ &= -{\boldsymbol \nabla} \frac{1}{|{\bf r} - {\bf r}'|} \cdot \bigl({\boldsymbol \nabla} \times {\bf J} ({\bf r}') \bigr) + {\bf J} ({\bf r}') \cdot \left({\boldsymbol \nabla} \times {\boldsymbol \nabla} \frac{1}{|{\bf r} - {\bf r}'|}\right) \label{Gr(5.47)} \end{align}

But \({\bf J}\) depends only on \({\bf r}'\) so \({\boldsymbol \nabla} \times {\bf J} ({\bf r}') = 0\), and since the curl of a gradient always vanishes, we obtain

The divergence of a magnetic field is always zero. Said equivalently: there are no magnetic charges.

We can play the same trick with the curl: \[ {\boldsymbol \nabla} \times {\bf B} ({\bf r}) = \frac{\mu_0}{4\pi} \int_{\cal V} d\tau' ~{\boldsymbol \nabla} \times \left(\frac{{\bf J}({\bf r}') \times ({\bf r} - {\bf r}')}{|{\bf r} - {\bf r}'|^3} \right) \label{Gr(5.49)} \] Do as above but now use product rule 8 \[ {\boldsymbol \nabla} \times ({\bf A} \times {\bf B}) = ({\bf B} \cdot {\boldsymbol \nabla}) {\bf A} - ({\bf A} \cdot {\boldsymbol \nabla}) {\bf B} + {\bf A} ({\boldsymbol \nabla} \cdot {\bf B}) - {\bf B} ({\boldsymbol \nabla} \cdot {\bf A}) \] and (dropping terms involving derivatives of \({\bf J}({\bf r}')\) with respect to \({\bf r}\), which vanish):

\begin{align} {\boldsymbol \nabla} \times \left(\frac{{\bf J}({\bf r}') \times ({\bf r} - {\bf r}')}{|{\bf r} - {\bf r}'|^3} \right) &= -{\boldsymbol \nabla} \times \left( {\bf J}({\bf r}') \times {\boldsymbol \nabla} \frac{1}{|{\bf r} - {\bf r}'|} \right) \nonumber \\ &= ({\bf J} ({\bf r}') \cdot {\boldsymbol \nabla}) {\boldsymbol \nabla} \frac{1}{|{\bf r} - {\bf r}'|} -{\bf J} ({\bf r}') \left({\boldsymbol \nabla} \cdot {\boldsymbol \nabla} \frac{1}{|{\bf r} - {\bf r}'|}\right) \label{Gr(5.50)} \end{align}

Last term: \[ -{\bf J} ({\bf r}') \left({\boldsymbol \nabla} \cdot {\boldsymbol \nabla} \frac{1}{|{\bf r} - {\bf r}'|}\right) = -{\bf J} ({\bf r}') \left({\boldsymbol \nabla}^2 \frac{1}{|{\bf r} - {\bf r}'|}\right) = -{\bf J} ({\bf r}') \left(-4\pi \delta^{(3)} ({\bf r} - {\bf r}') \right) \label{Gr(5.51)} \] where we have used Lap1or. This term thus integrates to \[ \frac{\mu_0}{4\pi} \int_{\cal V} d\tau' {\bf J} ({\bf r}') 4\pi \delta^{(3)} ({\bf r} - {\bf r}') = \mu_0 {\bf J}({\bf r}). \] To treat the other term, we use \[ ({\bf J} ({\bf r}') \cdot {\boldsymbol \nabla}) {\boldsymbol \nabla} \frac{1}{|{\bf r} - {\bf r}'|} = -({\bf J} ({\bf r}') \cdot {\boldsymbol \nabla}') {\boldsymbol \nabla} \frac{1}{|{\bf r} - {\bf r}'|} \] to rewrite the second part of the integral as \[ -\frac{\mu_0}{4\pi} {\boldsymbol \nabla} \int_{\cal V} d\tau' ({\bf J} ({\bf r}') \cdot {\boldsymbol \nabla}') \frac{1}{|{\bf r} - {\bf r}'|} = \frac{\mu_0}{4\pi} {\boldsymbol \nabla} \int_{\cal V} d\tau' \frac{{\boldsymbol \nabla}' \cdot {\bf J} ({\bf r}')}{|{\bf r} - {\bf r}'|} = 0 \] where in the second step we have used integration by parts using product rule 5 \[ {\boldsymbol \nabla} \cdot (f {\bf A}) = f ({\boldsymbol \nabla} \cdot {\bf A}) + {\bf A} \cdot ({\boldsymbol \nabla} f) \] (the surface term vanishes because we take \({\bf J} \rightarrow 0\) at infinity), and in the third step we have used the assumption of steady-state so \({\boldsymbol \nabla}' \cdot {\bf J} = 0\).

We thus obtain in total

(in differential form). Using Stokes' theorem, \[ \int_{\cal S} {\boldsymbol \nabla} \times {\bf B} \cdot d{\bf a} = \oint_{\cal P} {\bf B} \cdot d{\bf l} = \mu_0 \int_{\cal S} {\bf J} \cdot d{\bf a} \] so

where \(I_{enc}\) is the current enclosed in the {\bf amperian loop} \({\cal P}\) which defines the boundary of surface \({\cal S}\).

Sign ambiguity: resolved by right-hand rule as usual.

Ampère's law in magnetostatics takes a parallel role to Gauss's law in electrostatics.

\paragraph{Situations where Ampère's law can be useful:} i) infinite straight lines, ii) infinite planes, iii) infinite solenoids, iv) toroids.