# Pre-Quantum Electrodynamics
Author: Jean-Sébastien Caux

##### Divergence and Curl of \({\bf B}\) from Biot-Savart
ems.ms.dcB.BS

\[ {\bf B} ({\bf r}) = \frac{\mu_0}{4\pi} \int_{\cal V} d\tau' \frac{{\bf J}({\bf r}') \times ({\bf r} - {\bf r}')}{|{\bf r} - {\bf r}'|^3} \label{Gr(5.45)} \] 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 \ref{Gr(1.101)}) \(\frac{{\bf r} - {\bf r}'}{|{\bf r} - {\bf r}'|^3} = -{\boldsymbol \nabla} \frac{1}{|{\bf r} - {\bf r}'|}\). 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

\[ {\boldsymbol \nabla} \cdot {\bf B} = 0 \label{eq:DivBisZero} \]

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 \ref{Gr(1.102)}. 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

Ampère's law \[ {\boldsymbol \nabla} \times {\bf B} = \mu_0 {\bf J} \label{eq:Ampere} \]

(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

\[ \oint_{\cal P} {\bf B} \cdot d{\bf l} = \mu_0 I_{enc} \hspace{2cm} I_{enc} = \int_{\cal S} {\bf J} \cdot d{\bf a}. \label{Gr(5.55)} \]

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{Example 5.7:} same as Example 5.5, but now with Ampère. \paragraph{Solution:} by symmetry, \({\bf B}\) is circumferential and can only depend on \(s\). Then, choosing an amperian loop at a fixed radius \(s\), we get \[ \oint {\bf B} \cdot d{\bf l} = B 2\pi s = \mu_0 I ~~\Rightarrow~~ B = \frac{\mu_0 I}{2\pi s} \]

\paragraph{Example 5.8:} uniform surface current \({\bf K} = K \hat{\bf x}\) flowing in \(xy\) plane. \paragraph{Solution:} Biot-Savart: \({\bf B}\) must be perpendicular to \({\bf K}\). Intuition: \({\bf B}\) cannot have a component along \(\hat{\bf z}\). Right-hand rule: \({\bf B}\) along \(-\hat{\bf y}\) for \(z > 0\), and along \(\hat{\bf y}\) for \(z < 0\). Amperian loop of width \(l\) punching through surface: \[ \oint {\bf B} \cdot d{\bf l} = 2B l = \mu_0 I_{enc} = \mu_0 K l ~~\Rightarrow~~ {\bf B} = \left\{ \begin{array}{cc} \frac{\mu_0}{2} K \hat{\bf y}, & z < 0, \\ -\frac{\mu_0}{2} K \hat{\bf y}, & z > 0. \end{array} \right. \label{Gr(5.56)} \]

\paragraph{Example 5.9:} solenoid along \(\hat{\bf z}\), wire carrying current \(I\) doing \(n\) turns per unit length on cylinder of radius \(R\). \paragraph{Solution:} by symmetry, \({\bf B}\) must be along axis of solenoid. Outside: infinitely far away, \({\bf B}\) must vanish. But an amperian loop outside gives zero always, so \({\bf B}\) vanishes everywhere outside the solenoid. Amperian loop of length \(l\), half-inside and half-outside: \[ \oint {\bf B} \cdot d{\bf l} = Bl = \mu_0 I_{enc} = \mu_0 I n l ~~\Rightarrow~~ {\bf B} = \left\{ \begin{array}{cc} \mu_0 I n \hat{\bf z}, & s < R, \\ 0, & s > R \end{array} \right. \label{Gr(5.57)} \]

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

\paragraph{Example 5.10:} toroidal coil (no matter the shape, as long as it is rotationally symmetric). \paragraph{Solution:} magnetic field is circumferential everywhere. Outside coil, field again zero. Amperian loop half inside, half outside: \[ B 2\pi s = \mu_0 I_{enc} ~~\Rightarrow~~ {\bf B} = \left\{ \begin{array}{cc} \frac{\mu_0 N I}{2\pi s} \hat{\boldsymbol \phi}, & \mbox{inside coil}, \\ 0, & \mbox{outside} \end{array} \right. \label{Gr(5.58)} \]

Author: Jean-Sébastien Caux

Created: 2022-02-08 Tue 06:55

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