Pre-Quantum Electrodynamics
Ampère's Law in Magnetized Materialsemsm.msm.H.A
Nomenclature: as in electric case, we have bound currents, and everything else, which we call the free current. Total current:
\[ {\bf J} = {\bf J}_b + {\bf J}_f \tag{JbJf}\label{JbJf} \] Ampère's law: \[ \frac{1}{\mu_0} ({\boldsymbol \nabla} \times {\bf B}) = {\bf J} = {\bf J}_f + {\bf J}_b = {\bf J}_f + ({\boldsymbol \nabla} \times {\bf M}), \] so we can define
and rewrite Ampère's law as
or in integral form,
\({\bf H}\) in magnetostatics: parallel role to \({\bf D}\) in electrostatics. Allows us to rewrite Ampère's law in terms of free currents alone. Bound currents come along for the ride.
Example: copper rod
Consider a long copper rod of radius \({\bf R}\) carrying uniformly distributed free current \(I\).
Task: find \({\bf H}\) inside and outside the rod.
Solution: copper weakly diamagnetic: dipoles line up opposite the field. Bound currents antiparallel to \(I\) in bulk and parallel at surface. All currents longitudinal so \({\bf B}, {\bf M}, {\bf H}\) are circumferential. Apply integral form of Ampère's law with radius \(s < R\): \(H (2\pi s) = I_{f_{enc}} = I \frac{\pi s^2}{\pi R^2}\) so
\[ {\bf H} = \frac{I s}{2\pi R^2} \hat{\boldsymbol \varphi}, \hspace{5mm} s \leq R \tag{Hrod_in}\label{Hrod_in} \] Outside,
\[ {\bf H} = \frac{I}{2\pi s} \hat{\boldsymbol \varphi}, \hspace{5mm} s \geq R. \tag{Hrod_out}\label{Hrod_out} \] There, \({\bf M} = 0\) so \({\bf B} = \mu_0 {\bf H}\).
Energy in Linear Media
Need W_intAJ (to be proven later) giving the magnetic energy of a system of free currents: \[ W_{mag} = \frac{1}{2} \int_{\cal V} d\tau {\bf A} \cdot {\bf J}_f \] Similarly to the electric case, we can consider the presence of linear media. Then, work necessary to increase flux is (from EMF) \(\Delta W_{mag} = V_{ext} \Delta q = V I \Delta t = I \Delta \phi\) so \[ \Delta W_{mag} = I \Delta \phi \] In terms of current density: use \(\phi = \int_{\cal S} (\boldsymbol \nabla \times {\bf A}) \cdot d{\bf a} = \int_{\cal C} {\bf A} \cdot d{\bf s}\), move to volume currents: \[ \Delta W_{mag} = \int_{\cal V} d\tau {\bf J}_f \cdot \Delta {\bf A} = \int_{\cal V} d\tau ({\boldsymbol \nabla} \times {\bf H}) \cdot \Delta {\bf A} \] But \({\boldsymbol \nabla} \times (\Delta {\bf A}) = \Delta {\bf B}\) and
\begin{equation*} ({\boldsymbol \nabla} \times {\bf H}) \cdot \Delta {\bf A} = {\bf H} \cdot ({\boldsymbol \nabla} \times \Delta {\bf A}) - {\boldsymbol \nabla} \cdot (\Delta {\bf A} \times {\bf H}) = {\bf H} \cdot \Delta {\bf B} - {\boldsymbol \nabla} \cdot (\Delta {\bf A} \times {\bf H}) \end{equation*}Integrating, we get \[ \Delta W_{mag} = \int_{all~space} d\tau {\bf H} \cdot \Delta {\bf B} \] Case of linear isotopic homogeneous medium: \[ W_{mag} = \int_{all~space} d\tau \frac{1}{2} {\bf H} \cdot {\bf B} \]
Boundary conditions
Can rewrite BCs in terms of \({\bf H}\): from divH,
\begin{equation} H^{\perp}_{above} - H^{\perp}_{below} = -(M^{\perp}_{above} - M^{\perp}_{below}) \tag{HdiscM}\label{HdiscM} \end{equation}while curlHJf gives
\begin{equation} {\bf H}^{\parallel}_{above} - {\bf H}^{\parallel}_{below} = {\bf K}_f \times \hat{\bf n} \tag{Hdisc}\label{Hdisc} \end{equation}These are more useful than BCs on \({\bf B}\), Bdisc: \[ B^{\perp}_{above} = B^{\perp}_{below}. \label{Gr(6.26)} \] and \[ {\bf B}^{\parallel}_{above} - {\bf B}^{\parallel}_{below} = \mu_0 K \times \hat{\bf n} \label{Gr(6.27)} \]
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Created: 2024-02-27 Tue 10:31