Pre-Quantum Electrodynamics
Curl of \({\bf B}\) from Biot-Savart; Ampère's Lawems.ms.dcB.c
- Gr 5.3.2
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 curl_xprod 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}') \left({\boldsymbol \nabla} \cdot {\boldsymbol \nabla} \frac{1}{|{\bf r} - {\bf r}'|}\right) + ({\bf J} ({\bf r}') \cdot {\boldsymbol \nabla}) {\boldsymbol \nabla} \frac{1}{|{\bf r} - {\bf r}'|} \end{align*}The first term can be reworked into
\begin{align*} -{\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) \end{align*}where we have used Lap1or. Back in Biot-Savart, 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 that for a symmetric function of \({\bf r} - {\bf r}'\), changing from \(\nabla\) to \(\nabla'\) simply changes the sign, i.e. \[ ({\bf J} ({\bf r}') \cdot {\boldsymbol \nabla}) {\boldsymbol \nabla} \frac{1}{|{\bf r} - {\bf r}'|} = -({\bf J} ({\bf r}') \cdot {\boldsymbol \nabla}') {\boldsymbol \nabla} \frac{1}{|{\bf r} - {\bf r}'|} \] This allows us 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 div_prod (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
Ampère's law
\[ {\boldsymbol \nabla} \times {\bf B} = \mu_0 {\bf J} \tag{Amp}\label{Amp} \]
(in differential form). Using Stokes' theorem, \[ \int_{\cal S} d{\bf a} \cdot {\boldsymbol \nabla} \times {\bf B} = \oint_{\cal P} d{\bf l} \cdot {\bf B} = \mu_0 \int_{\cal S} d{\bf a} \cdot {\bf J} \] yielding
Ampère's law (integral form)
\[ \oint_{\cal P} d{\bf l} \cdot {\bf B} = \mu_0 I_{enc} \hspace{2cm} I_{enc} = \int_{\cal S} d{\bf a} \cdot {\bf J}. \label{Gr(5.55)} \]
where \(I_{enc}\) is the current enclosed in the 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.
Example: infinite wire
Same as direct integration we did before, but let's now use Ampère.
Solution: by symmetry, \({\bf B}\) is circumferential and can only depend on \(r\). Then, choosing an amperian loop at a fixed radius \(r\), we get \[ \oint d{\bf l} \cdot {\bf B} = B 2\pi r = \mu_0 I ~~\Rightarrow~~ B = \frac{\mu_0 I}{2\pi r} \]
Example: uniform sheet current
Consider an infinite sheet with uniform surface current \({\bf K} = K \hat{\bf x}\) flowing in \(xy\) plane.
Task: calculate \({\bf B}\).
Solution: from Biot-Savart BiotSavart_s, \({\bf B}\) 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: \[ \oint {\bf B} \cdot d{\bf l} = 2B l = \mu_0 I_{enc} = \mu_0 K l ~~\Rightarrow~~ {\bf B} = \left\{ \begin{array}{cc} \frac{\mu_0}{2} K \hat{\bf y}, & z < 0, \\ -\frac{\mu_0}{2} K \hat{\bf y}, & z > 0. \end{array} \right. \label{Gr(5.56)} \]
Example: solenoid
Consider an infinite solenoid along \(\hat{\bf z}\), made of a wire carrying current \(I\) doing \(n\) turns per unit length on a cylinder of radius \(R\).
Task: calculate \({\bf B}\).
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:
\[ \oint d{\bf l} \cdot {\bf B} = Bl = \mu_0 I_{enc} = \mu_0 I n l ~~\Rightarrow~~ {\bf B} = \left\{ \begin{array}{cc} \mu_0 I n ~\hat{\bf z}, & r < R, \\ 0, & r > R \end{array} \right. \tag{Bsol}\label{Bsol} \]
N.B.: the situations in which Ampère's law can be useful are those involving some form of symmetry, for example i) infinite straight lines, ii) infinite planes, iii) infinite solenoids, iv) toroids.
Example: toroidal coil
Consider a toroidal coil (it doesn't matter what the cross-sectional shape, as long as it is rotationally symmetric).
Task: calculate \({\bf B}\).
Solution: by rotational symmetry, the magnetic field is circumferential everywhere. Outside the coil, the field must again be zero.
Consider now an Amperian loop which is just a circle at radius \(r\):
\[ B 2\pi r = \mu_0 I_{enc} ~~\Rightarrow~~ {\bf B} = \left\{ \begin{array}{cc} \frac{\mu_0 N I}{2\pi r} \hat{\boldsymbol \varphi}, & \mbox{inside coil}, \\ 0, & \mbox{outside} \end{array} \right. \tag{Btor}\label{Btor} \]
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Created: 2024-02-27 Tue 10:31