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 {\bf free current}. Total current: \[ {\bf J} = {\bf J}_b + {\bf J}_f \label{Gr(6.17)} \] 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
\[ {\bf H} \equiv \frac{1}{\mu_0} {\bf B} - {\bf M} \label{Gr(6.18)} \]
and rewrite Ampère's law as
\[ {\boldsymbol \nabla} \times {\bf H} = {\bf J}_f \label{Gr(6.19)} \]
or in integral form,
\[ \oint {\bf H} \cdot d{\bf l} = I_{f_{enc}} \label{Gr(6.20)} \]
\({\bf H}\) 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.
\paragraph{Example 6.2:} long copper rod radius \({\bf R}\) carries uniformly distributed free current \(I\). Find \({\bf H}\) inside and outside rod. \paragraph{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 \[ {\bf H} = \frac{I s}{2\pi R^2} \hat{\boldsymbol \varphi}, \hspace{5mm} s \leq R \label{Gr(6.21)} \] Outside, \[ {\bf H} = \frac{I}{2\pi s} \hat{\boldsymbol \varphi}, \hspace{5mm} s \geq R. \label{Gr(6.22)} \] There, \({\bf M} = 0\) so \({\bf B} = \mu_0 {\bf H}\).
A Deceptive Parallel
Similarly to electric case: cannot assume that \({\bf H}\) is like \({\bf B}\). \({\bf H}\) might have a divergence, \[ {\boldsymbol \nabla} \cdot {\bf H} = -{\boldsymbol \nabla} \cdot {\bf M} \label{Gr(6.23)} \]
Energy in Linear Media
Recall \ref{Gr(7.31)}, magnetic energy of 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 \[ ({\boldsymbol ∇} × {\bf H}) ⋅ Δ {\bf A} = {\bf H} ⋅ ({\boldsymbol ∇} × Δ {\bf A})
- {\boldsymbol ∇} ⋅ (Δ {\bf A} × {\bf H}) = {\bf H} ⋅ Δ {\bf B} - {\boldsymbol ∇} ⋅ (Δ {\bf A} × {\bf H})
\] Integrating, we get \[ \Delta W_{mag} = \int_{all~space} d\tau {\bf H} \cdot \Delta {\bf B} \] Case of linear isotopic homogeneous medium: \[ W_{mag} = \int_{all~space} d\tau \frac{1}{2} {\bf H} \cdot {\bf B} \]
\subsubsection*{Boundary conditions} Can rewrite BCs in terms of \({\bf H}\): from \ref{Gr(6.23)}, \[ H^{\perp}_{above} - H^{\perp}_{below} = -(M^{\perp}_{above} - M^{\perp}_{below}) \label{Gr(6.24)} \] while \ref{Gr(6.19)} gives \[ {\bf H}^{\parallel}_{above} - {\bf H}^{\parallel}_{below} = {\bf K}_f \times \hat{\bf n} \label{Gr(6.25)} \] These are more useful than BCs on \({\bf B}\), \ref{Gr(5.72)} and \ref{Gr(5.73)}: \[ B^{\perp}_{above} = B^{\perp}_{below}. \label{Gr(6.26)} \] 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: 2022-02-21 Mon 20:41