# Pre-Quantum Electrodynamics
Author: Jean-Sébastien Caux

#### Energy in Magnetic Fields
emd.Fl.e

Work per unit time in current loop: \[ \frac{dW}{dt} = -{\cal E} I = L I \frac{dI}{dt} \] Start from zero current, integrate in time: \[ W = \frac{1}{2} L I^2 \label{Gr(7.29)} \] Nicer way (generalizable to surface and volume currents): from (\ref{Gr(7.25)}), flux through loop is \(\Phi = L I\). But \[ \Phi = \int_{\cal S} {\bf B} \cdot d{\bf a} = \int_{\cal S} ({\boldsymbol \nabla} \times {\bf A}) \cdot d{\bf a} = \oint_{\cal P} {\bf A} \cdot d{\bf l}, \] so \[ LI = \oint {\bf A} \cdot d{\bf l} \] and \[ W = \frac{1}{2} I \oint {\bf A} \cdot d{\bf l} = \frac{1}{2} \oint ({\bf A} \cdot {\bf I}) dl \label{Gr(7.30)} \] Generalization to volume currents: \[ W = \frac{1}{2} \int_{\cal V} ({\bf A} \cdot {\bf J}) d\tau \label{Gr(7.31)} \] Even better: use Ampère, \({\boldsymbol \nabla} \times {\bf B} = \mu_0 {\bf J}\): \[ W = \frac{1}{2\mu_0} \int_{\cal V} {\bf A} \cdot ({\boldsymbol \nabla} \times {\bf B}) d\tau \label{Gr(7.32)} \] Integrate by parts using product rule 6: \[ {\boldsymbol ∇} ⋅ ({\bf A} × {\bf B}) = {\bf B} ⋅ ({\boldsymbol ∇} × {\bf A})

- {\bf A} ⋅ ({\boldsymbol ∇} × {\bf B}),

\] so \[ {\bf A} ⋅ ({\boldsymbol ∇} × {\bf B}) = {\bf B} ⋅ {\bf B}

- {\boldsymbol ∇} ⋅ ({\bf A} × {\bf B}).

\] Then, \[ W = \frac{1}{2\mu_0} \left[ \int_{\cal V} d\tau B^2 - \int_{\cal V} d\tau {\boldsymbol \nabla} \cdot ({\bf A} \times {\bf B}) \right] = \frac{1}{2\mu_0} \left[ \int_{\cal V} d\tau B^2 - \oint_{\cal S} d{\bf a} \cdot ({\bf A} \times {\bf B}) \right] \label{Gr(7.33)} \] We can integrate over all space: after neglecting boundary terms (assuming fields fall to zero at infinity), we are left with

\[ W_{mag} = \frac{1}{2\mu_0} \int d\tau B^2 \label{Gr(7.34)} \]

Summary: energy in electric and magnetic fields:

\begin{align} W_{elec} = \frac{1}{2} \int d\tau V\rho = \frac{\varepsilon_0}{2} \int d\tau E^2, \hspace{2cm} \mbox{(2.43 and 2.45)}, \\ W_{mag} = \frac{1}{2} \int d\tau ({\bf A} \cdot {\bf J}) = \frac{1}{2\mu_0} \int d\tau B^2, \hspace{2cm} \mbox{(7.31 and 7.34)} \end{align}

\paragraph{Example 7.13:} coaxial cable (inner cylinder radius \(a\), outer \(b\)) carries current \(I\). Find energy stored in section of length \(l\). \paragraph{Solution:} from Ampère, \[ {\bf B} = \frac{\mu_0 I}{2\pi s} \hat{\boldsymbol \phi}, \hspace{1cm} a < s < b, \hspace{1cm} {\bf B} = 0, \hspace{1cm} s < a ~\mbox{or}~ s > b. \] Energy is thus \[ W_{mag} = \frac{1}{2\mu_0} \int_0^{2\pi} d\phi \int_0^l dz \int_a^b s ds \left(\frac{\mu_0 I}{2\pi s}\right)^2 = \frac{\mu_0 I^2 l}{4\pi} \ln \frac{b}{a}. \]

Note: gives easy way to find inductance, since \(W = \frac{1}{2} L I^2\).

Author: Jean-Sébastien Caux

Created: 2022-02-08 Tue 06:55

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