\begin{problem*}{24.25} % Poynting vector, power <-> EM fields
The filament of an incandescent lamp has a $150\U{\Ohm}$ resistance
and carries a direct current of $1.00\U{A}$.  The filament is
$8.00\U{cm}$ long and $0.900\U{mm}$ in radius.  \Part{a} Calculate
the Poynting vector at the surface of the filament, associated with
the static electric field producing the current and the curret's
static magnetic field.  \Part{b} Find the magnitude of the static
electric and magnetic fields at the surface of the filament.
\end{problem*}

\begin{solution}
\Part{a}
The hot resistor will be radiating heat, and none of the electric or
magnetic fields change with time, so we expect a constant Poynting
vector of magnitude
\begin{equation}
  S = \frac{P}{A} = \frac{I^2R}{2\pi r L}
    = \ans{332\U{kW/m$^2$}} \;.
\end{equation}
This Poynting vector will always point away from the wire (in the
direction the radiation is going).

\Part{b}
The electric field is given by Ohm's law.
\begin{align}
  V &= I R \\
  E &= \frac{V}{L} = \frac{IR}{L}
    = \frac{1.00\U{A}\cdot150\U{\Ohm}}{8.00\E{-2}\U{m}}
    = \ans{1875\U{V/m}} \;.
\end{align}
The magnetic field from a long, straight wire is
\begin{equation}
  B = \ans{\frac{I}{2\pi r}} \;,
\end{equation}
so the magnetic field at the surface of the wire is
\begin{equation}
  B = 177\U{T}
\end{equation}
The electric field is along the wire, and the magnetic field is
perpendicular to the current, so the Poynting vector points directly
out (perpendicular to the wire's surface) and has a magnitude
\begin{equation}
  S = EB\sin(90\dg) = EB = \frac{IR}{L}\cdot\frac{I}{2\pi r}
    = \frac{I^2R}{2\pi r L} \;,
\end{equation}
which is the same expression we found in \Part{a}.
\end{solution}
