[NOTES/EM-09005]-Energy Stored in Magnetic Field

The energy stored in a carrying current coil is computed using the rate of work done \begin{equation} \dd[W]{t} = \mathcal E I = \dd[\Phi] I = I L \dd[I]{t}. \end{equation} Here \(\Phi=LI\) is just the flux linked with the coil, and is given by \begin{equation} LI =\Phi = \iint_S \vec{B}\cdot \hat{n} dS = \oint_\text{loop} \vec{A}\cdot \overrightarrow{d\ell} \end{equation} where the magnetic field is written in terms of vector potential, \(\vec{B}=\nabla \times \vec{A}\), and Stokes theorem has been used. Thus the work done in establishing a current \(I\) is given by \begin{equation} W =\frac{1}{2} L I^2 = \frac{1}{2} I \oint_\Gamma \vec{A}\cdot \overrightarrow{d\ell} \end{equation} In a circuit the direction of the current \(I\) coincides with the direction of the line element \(\overrightarrow{d\ell}\). Using \(\vec{I}\) to denote the vector representing the current, we have \(I\overrightarrow{d\ell} = \vec{I} d\ell\). Therefore, % \begin{align*} W =& \frac{1}{2}\int(\vec{A}\cdot\vec{I})dl. \end{align*} The above expression of the energy can be generalized to volume current by using the correspondence \(\vec{I}d\ell \leftrightarrow \vec{j} dV\). Therefore, we get \begin{eqnarray} W &= & \frac{1}{2}\iiint_V \vec{A}\cdot\vec{J}\,dl \rightarrow \frac{1}{2} \iiint_{-\infty}^\infty \vec{A}\cdot\vec{J}~d^3x \end{eqnarray} The integral in the first step is over the volume \(V\), (assumed to be a bounded volume) , where the current \(\vec{J}\) is non zero. In the next step the range of integration has been extended to all space because $\vec{J}=0$ outside volume $V$. \begin{eqnarray}\nonumber W &=& \frac{1}{2\mu_0} \int\bar{A}\cdot(\bar{\nabla}\times\bar{B})dV\\ && \qquad \qquad {\HighLight{using  \nabla\cdot(\bar{A}\times\bar{B}) = \bar{B}\cdot(\bar{\nabla}\times\bar{A}) -\bar{A}\cdot(\bar{\nabla}\times\bar{B})}}\\\nonumber &=&\frac{1}{2\mu_0} \int \vec{B}\cdot(\bar{\nabla}\times\bar{A})dV - \int \nabla\cdot(\bar{A}\times\bar{B})dV\\\nonumber &=&\frac{1}{2\mu_0} \int \vec{B}\cdot(\bar{\nabla}\times\bar{A})dV + \text{All space}\\\nonumber &=&\frac{1}{2\mu_0} \int \vec{B}\cdot(\bar{\nabla}\times\bar{A})dV + \lim \int_{V}\nabla\cdot(\bar{A}\times\bar{B})dV\\ &=&\frac{1}{2\mu_0} \int \vec{B}\cdot(\bar{\nabla}\times\bar{A})dV + \lim_{R\to\infty} \int(\vec{a}\times\vec{B})\cdot\hat{n} \, dS \end{eqnarray} The surface term vanishes as \(R\to\infty\) because \(\vec{A}~\text{and}~\vec{B} \rightarrow0~\text{as}~R\to\infty\) So the energy stored in building up magnetic field (or setting up a current) is \begin{equation}\boxed{ W = \frac{1}{2\mu_0} \iiint_{-\infty}^\infty (\vec{B}\cdot\vec{B})dV .} \end{equation}