#### 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 PLI, 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:

- Gr (7.31)

\[ W = \frac{1}{2} \int_{\cal V} d\tau ~({\bf A} \cdot {\bf J}) \tag{W_intAJ}\label{W_intAJ} \] Even better: use Ampère, \({\boldsymbol \nabla} \times {\bf B} = \mu_0 {\bf J}\): \[ W = \frac{1}{2\mu_0} \int_{\cal V} d\tau ~{\bf A} \cdot ({\boldsymbol \nabla} \times {\bf B}) \label{Gr(7.32)} \] Integrate by parts using product rule 6: \[ {\boldsymbol \nabla} \cdot ({\bf A} \times {\bf B}) = {\bf B} \cdot ({\boldsymbol \nabla} \times {\bf A}) - {\bf A} \cdot ({\boldsymbol \nabla} \times {\bf B}), \] so \[ {\bf A} \cdot ({\boldsymbol \nabla} \times {\bf B}) = {\bf B} \cdot {\bf B} - {\boldsymbol \nabla} \cdot ({\bf A} \times {\bf B}). \] 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

- Gr (7.34)

\[ W_{mag} = \frac{1}{2\mu_0} \int d\tau B^2 \tag{W_intBsq}\label{W_intBsq} \]

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, \\ W_{mag} &= \frac{1}{2} \int d\tau ~({\bf A} \cdot {\bf J}) &= \frac{1}{2\mu_0} \int d\tau ~B^2, \end{align}

which are equations W_vcd, W_intEsq, W_intAJ and W_intBsq.

Example: energy in coaxial cable

Consider a coaxial cable with inner cylinder radius \(a\), outer \(b\), carrying current \(I\).

Task: find the energy stored in a section of length \(l\).

Solution: from Ampère, \[ {\bf B} = \frac{\mu_0 I}{2\pi s} \hat{\boldsymbol \varphi}, \hspace{1cm} a < s < b, \hspace{1cm} {\bf B} = 0, \hspace{1cm} s < a ~\mbox{or}~ s > b. \] Energy is thus \[ W_{mag} = \frac{1}{2\mu_0} \int_0^{2\pi} d\varphi \int_0^l dz \int_a^b s ds \left(\frac{\mu_0 I}{2\pi s}\right)^2 = \frac{\mu_0 I^2 l}{4\pi} \ln \frac{b}{a}. \]

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

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Author: Jean-Sébastien Caux

Created: 2022-03-24 Thu 08:42