Page 184 - Electromagnetics

Joule’s law. In § 2.9.5 we showed that when a moving charge interacts with an electric
field in a volume region V , energy is transferred between the field and the charge. If the
source of that energy is outside V , the energy is carried into V as an energy flux over the
boundary surface S. The energy balance described by Poynting’s theorem (3.299)also
holds for static fields supported by steady currents: we must simply recognize that we
have no time-rate of change of stored energy. Thus

− J · E dV = (E × H) · dS. (3.180)
V S
The term

P =− J · E dV (3.181)
V
describes the rate at which energy is supplied to the fields by the current within V ;we
have P > 0 if there are sources within V that result in energy transferred to the fields,
and P < 0 if there is energy transferred to the currents. The latter case occurs when
there are conducting materials in V . Within these conductors

P =− σE · E dV. (3.182)
V
Here P < 0; energy is transferred from the fields to the currents, and from the currents
into heat (i.e., into lattice vibrations via collisions). Equation (3.182) is called Joule’s
law, and the transfer of energy from the fields into heat is Joule heating. Joule’s law is
the power relationship for a conducting material.
An important example involves a straight section of conducting wire having circular
cross-section. Assume a total current I is uniformly distributed over the cross-section
of the wire, and that the wire is centered on the z-axis and extends between the planes
z = 0, L. Let the potential difference between the ends be V . Using (3.169)we see that
at the surface of the wire
I V
ˆ
H = φ , E = ˆ z .
2πa L
The corresponding Poynting flux E × H is −ˆρ-directed, implying that energy flows into
wire volume through the curved side surface. We can verify (3.180):
L 2π a I V
− J · E dV = ˆ z 2 · ˆ z ρ dρ dφ dz =−IV,
V 0 0 0 πa L
2π L IV
(E × H) · dS = −ˆρ · ˆρadφ dz =−IV.
S 0 0 2πaL

Stored magnetic energy. We have shown that the energy stored in a static charge
distribution may be regarded as the “assembly energy” required to bring charges from
infinity against the Coulomb force. By proceeding very slowly with this assembly, we are
able to avoid any complications resulting from the motion of the charges.
Similarly, we may equate the energy stored in a steady current distribution to the en-
6
ergy required for its assembly from current filaments brought in from infinity. However,
the calculation of assembly energy is more complicated in this case: moving a current
6 Recall that a flux tube of a vector field is bounded by streamlines of the field. A current filament is a
flux tube of current having vanishingly small, but nonzero, cross-section.

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