[NOTES/EM-10002]-Poynting Theorem--Examples

Energy stored in fields --- Examples

 Maxwell's Equation \begin{align*} \bar{\nabla}\cdot\bar{E} = & \frac{\rho}{\epsilon_0} &\bar{\nabla}\cdot\bar{B}=0&\\ \bar{\nabla}\times\bar{E}=&-\frac{\partial\vec{B}}{\partial t} &\vec{\nabla}\times\vec{B}=\mu_0\left(\vec{j}+\epsilon_0 \frac{\partial\vec{E}}{\partial t}\right)\\ &&\uparrow \vec{j}D \end{align*} Displacement current density $$ = \epsilon_0 ~\frac{\partial\vec{E}}{\partial t} $$ Poynting theorem $$ \frac{dW}{dt} + \frac{d}{dt} \iiint_V U d^3r + \iint_S \vec{\nabla}\cdot\vec{N}~dS = 0 $$ $U=$ energy density $$ =\frac{1}{2}\left(\epsilon_0\vec{E}\cdot{E}+\frac{1}{2\mu_0} \vec{B}\cdot\vec{B}\right) $$ $\vec{N}=$ Poynting vector $=\frac{1}{2\mu_0}(\vec{E}\times\vec{B})$ gives flow of energy per unit area per sec.

1.Magnetic energy of a solenoid

Energy of a solenoid

The energy of a solenoid carrying a current \(I\) and having self inductance \(\mathcal L\) is given by \[ \mathscr{E} = \frac{1}{2} \mathcal L I ^2.\] The magnetic field along the axis of a long solenoid is $\vec{B}=\mu_0In$, where $n$ is the number of turns per unit length of the solenoid. The magnetic field is zero outside the solenoid. The energy stored in the magnetic field per unit volume is $\frac{1}{2\mu_0} ~\vec{B}\cdot\vec{B}$Since the field outside the solenoid is zero, $\vec{B}=0$, and inside it is a constant, the energy stored in the magnetic field is\\ \begin{eqnarray} \text{Energy stored} &=& \frac{1}{2\mu_0}\vec{B}\cdot\vec{B} \times \text{volume of solenoid} \\ &=& \left(\frac{1}{2\mu_0}\right)(\mu_0In)^2\cdot(\pi a^2)L\\ &=&\frac{1}{2}\mu_0I^2n^2LA \end{eqnarray} Compare with $\frac{1}{2}\mathcal{L} I^2$, we get the self inductance \begin{equation} \boxed{\mathcal{L} = \mu_0n^2LA } \end{equation} This expression agrees with the known result for the self inductance of a long solenoid. This suggests that the energy stored in the solenoid can be consistently interpreted as being associated with the magnetic field due the solenoid.

2.Energy of a charged capacitor 

Consider a parallel plate capacitor, of capacitance \(C\). If the capacitor plates are charged to a potential \(V\), \[ \text{the energy stored in the capacitor}=\frac{1}{2} C V^2 .\] We will now show that this expression coincides with the energy stored in the electric field between the plates of the capacitor. We assume that the distance between the plate is small compared with the plate's dimensions. Therefore, the electric field will be assumed to uniform between the plates. Also the field out side the region between the plates will be neglected. When the capacitor has charge \(q\) on the plates, the potential difference is given by $V=q/C$. This corresponds to an electric field between the plates \(=(V/d)\) where \(d\) is distance between plates. The energy stored in the electric field per unit volume being $\frac{1}{2} \epsilon_0\vec{E}^2 $, we get \begin{eqnarray}\label{EQ01} \text{Energy stored in electric field } &=& \frac{1}{2} \epsilon_0\vec{E}^2 \times \text{Volume}\\ &=& \frac{1}{2} \Big(\frac{V}{d}\Big)^2 Ad \end{eqnarray} where \(A\) is the are of plates. Substituting \(q/C\) for the potential difference \(V\), the expression \eqRef{EQ01} becomes \begin{align} \left(\frac{1}{2}\epsilon_0\vec{E}^2\right)\times ( Ad) = \frac{1}{2}\Big(\frac{\epsilon_0 A}{d}\Big) V^2.\label{EQ02} \end{align} Now recall that the capacitance is \(C= \epsilon_0A/d\), and that the capacitance of a parallel plate capacitor is $ C = (\epsilon_0A/d)$. Comparing {EQ02} with $\frac{1}{2} CV^2$, we see that the expression {EQ02} coincides with \(\frac{1}{2} C V^2\) for the energy of the capacitor.

3.Ampere's law and displacement current

Charging of a capacitor

If the charge on the plates is $q(t)$ at time $t$, then the charging current is $I=\frac{dq}{dt}$. Also $\vec{E}$ and $\frac{d\vec{E}}{dt}$ are non-zero between plates. We calculate displacement current density \begin{align*} J_d = \epsilon_0~\frac{\partial E}{\partial t} = \epsilon_0~\frac{\partial}{\partial t}\left(\frac{V}{d}\right) = \epsilon_0 ~\frac{1}{dC}\cdot\frac{dq}{dt} \end{align*} where \\ $V=$ potential difference at time $t$, $V=q/C$\\ $C=$ capacitance $=\frac{\epsilon_0A}{a}$\\ $A=$ area of plates\\ $d=$ distance between plates.\\ Now consider an Amperian loop \(\gamma\) as shown in the figure. The time varying electric field produces a circulating magnetic field as shown in \Figref{??}. It will now be verified that the circulation of the magnetic field around the loop \(\gamma\) is equal to \(\mu_0 I\) in accordance with the Ampere's law. The magnetic field due to time varying electric field is given by the Maxwell's equation \begin{equation} {\nabla}\times \vec{B} =\mu_0\vec{j}+\epsilon_0\mu_0~ \frac{\partial\vec{E}}{\partial t} \end{equation} Consider the line integral \begin{align*} \int_r\vec{B}\cdot\overrightarrow{dl} =& \iint_S \bar{\nabla}\times \bar{B}\cdot\overrightarrow{dS}\\ =& \iint\left(\mu_0\vec{j}+\epsilon_0\mu_0~ \frac{\partial\vec{E}}{\partial t}\right)\overrightarrow{dS}\qquad \qquad \therefore \vec{j}=0,~\text{between the plates}\\ =& \mu_0\left(\frac{\epsilon_0}{dC}~\frac{dq}{dt}\right)A \qquad\qquad\frac{\partial\bar{E}}{\partial t}~\text{$\parallel$ to the normal}\\ =& \mu_0I,\quad \because I=\frac{dq}{dt} \end{align*} We have used $C=\epsilon_0 A/d$. This completes the demonstration that the displacement current restores the Ampere's law for time varying situations.