Poynting’s theorem in an Ohmic wire
A constant and uniformly distributed current density J = σE flows along an infinite wire or radius
a and conductivity σ.
a) Calculate Poynting’s vector S = ǫ0 c2 E × B and discuss Poynting’s theorem, i.e. energy conservation.
b) Show that S′ = ΦJ, with Φ the electrostatic potential, is also a suitable choice for S (hint:
substitute E = −∇Φ in S and manipulate the result).
1
Solution
a) For symmetry reasons, the magnetic field is azimuthal and depends only on the radial coordinate
r. Applyng Ampère’s law to a circle along the axis for r < a yields
2πrB = µ0 πr2 J ,
(1)
to find
µ0
rσE φ̂ .
2
(2)
σ
1
E × B = −r E 2 .
µ0
2
(3)
B=
Thus, at a distance r form the axis
S=
S(r)
S′ (z)
r
S′ (z + h)
z
S(r)
h
The energy flux through the surface of a cylinder of radius r and
length h and coaxial to the wire (see figure) is thus
Z
ΦS = dΣS · n = −2πrhS = −πr2 hσE 2 .
(4)
The energy flows inwards, as energy has to be dissipated inside the
volume by Joule heating. In fact
Z
−ΦS = dV J · E = πr2 hJE = πr2 hσE 2 ,
(5)
which satisfies Poynting’s theorem.
Note that if we take r > a, we find B = µ0 Ja2 /2r and S = −a2 E 2 r/2r2 , so that the energy flux
is indipendent of r and equals the total dissipated power:
ΦS = −πa2 hσE 2 .
(6)
b) We must show that S is equal to S′ minus the curl of any vector function. Let us substitute
E = −∇Φ in the expression for S:
S=
1
1
E × B = − ∇Φ × B .
µ0
µ0
(7)
Now we use the vector identity
∇ × (ΦB) = ∇Φ × B + Φ∇ × B = ∇Φ × B + Φ(µ0 J) .
(8)
Taking ∇Φ × B from (7) and inserting it into (8), we obtain
S = ΦJ −
1
∇ × (ΦB) .
µ0
(9)
The last term is the curl of ΦB, thus we may redefine Poynting’s vector as
S′ = ΦJ .
2
(10)
To show that S′ is equivalent to S, we compute its flux through the same cylinder as above. Now S′
is parallel to the axis, hence only the base surfaces contribute to the flux:
Φ(S′ ) = πa2 [−S ′ (z) + S ′ (z + h)] = πa2 J[Φ(z + h) − Φ(z)] .
(11)
since Φ = −Ez, we obtain
Φ(S′ ) = πa2 [−S ′ (z) + S ′ (z + h)] = −πa2 hJE ,
which again gives the dissipated power into the volume (with a minus sign).
3
(12)