Energy in the Electromagnetic Field
Electromagnetic energy storage\begin{align}
U_E &= \tfrac{1}{2}CV^2
= \tfrac{1}{2}\eps\int E^2\,d\tau \\[4pt]
U_B &= \tfrac{1}{2}LI^2
= \tfrac{1}{2\muz}\int B^2\,d\tau
\end{align}
Energy densities:
\begin{equation}
u_E = \frac{\eps}{2}E^2,\quad
u_B = \frac{B^2}{2\muz},\quad
u = \frac{1}{2}\!\left(\eps E^2+\frac{B^2}{\muz}\right)
\end{equation}
Poynting's Theorem — Derivation
Use M4 to eliminate \(\vec{J}\) from \(\vec{J}\cdot\vec{E}\), then apply the
vector identity
\(\vec{E}\cdot(\nabla\times\vec{B}) =
\vec{B}\cdot(\nabla\times\vec{E}) - \nabla\cdot(\vec{E}\times\vec{B})\)
and M3:
Poynting's Theorem\begin{equation}
\frac{\partial u}{\partial t} + \nabla\cdot\vec{S} = -\vec{J}\cdot\vec{E}
\end{equation}
Poynting vector (energy flux, W/m\(^2\)):
\begin{equation}
\boxed{\vec{S} = \frac{1}{\muz}\vec{E}\times\vec{B}}
\end{equation}
Integral form (energy conservation):
\begin{equation}
-\frac{dU_\text{em}}{dt}
= \oint\vec{S}\cdot d\vec{a} + \int\vec{J}\cdot\vec{E}\,d\tau
\end{equation}
What Poynting's theorem saysThe rate of decrease of EM energy in a volume equals the power flowing out
through its surface (via \(\vec{S}\)) plus the power delivered to charges
(via \(\vec{J}\cdot\vec{E}\)). Energy propagates through the field,
not along the wires.
Example 3.8 Poynting vector — resistive wireWire: radius \(a\), length \(L\), resistivity \(\eta\), current \(I\).
\(E_z = \eta I/\pi a^2\) (along wire);
\(B_\phi = \muz I/2\pi a\) (circumferential).
\(\vec{S} = (\vec{E}\times\vec{B})/\muz\) points radially inward.
Power into wire: \(P = |\vec{S}|\cdot 2\pi aL = I^2R\).
\[\boxed{P = I^2R}\quad\checkmark\]
Joule heating is supplied by energy flowing inward from the surrounding field.
Example 3.9 Energy in a solenoidSolenoid: \(n\), \(R\), \(l\), current \(I\).
\(u_B = \muz n^2I^2/2\); \(U = u_B\cdot\pi R^2 l = \frac{1}{2}LI^2\). ✓
\[\boxed{U_B = \frac{1}{2}LI^2}\]
Example 3.10 Poynting vector — charging capacitorCircular capacitor (radius \(R\), separation \(d\)), voltage \(V(t)\).
\(E=V/d\); \(B=(\muz\eps R/2d)\,dV/dt\) at edge (\(s=R\)).
Power flowing in: \(P = S\cdot 2\pi Rd = \frac{d}{dt}(\frac{1}{2}CV^2)\).
\[\boxed{P = \frac{d}{dt}\!\left(\tfrac{1}{2}CV^2\right)}\quad\checkmark\]
Concept Checks — Energy, Poynting & Momentum
Concept check.Rectangular loop (\(a\times b\), resistance \(R\)), rotating at \(\omega\) in field \(B\). Maximum induced current:
Show answer
\(\mathcal{E}_\text{max}=Bab\omega\); \(I_\text{max}=Bab\omega/R\).
