ENERGY CONSERVATION IN ELECTROMAGNETIC FIELDS
We now look at energy conservation in electromagnetic fields. We begin by considering
the work done on moving charges by the fields. We have:
 
  

dW  F  dr  q  E  v  B   dr
Then
   dr
 
dW
  

 P  q E  v  B 
 qE  v  q  v  B   v
dt
dt
 

But v  B is perpendicular to v . Hence

 v  B   v  0


Thus static B fields do no work! Does this mean that B fields never do work? No. A time


dependent B field will produce an electric field (by Faraday’s law) and the E field will
generally do work. If we now have a collection of charges we get:
 
P   q i E  vi
i
In the continuum limit this becomes:
 
P   E  J dvol
(Recall that qv → Idl → kdA → Jdvol as the number of dimensions increases.) Hence the
work/sec done by the fields in a volume V is:
 
P   E  J dvol
vol
We now use the Maxwell equation:


  
   D
E
H  J  ,
 J  H 
t
t
to get

  
 
 D 

 E  J dvol   E    H 
 dvol
t 


vol
vol
But
  
     
   E  H     E   H  E    H 

  
 D 
  
H    E      E  H   E 
 dvol
t 

vol
vol


   B

 D 
 
    H 
E
 dvol    E  H   dS
t
t 
 
surf
 
  E  J dvol  
vol
We know that

E



D
dvol
t
is the rate at which energy is going into the electric field. Similarly:
 
H



B
dvol
t
is the rate at which energy is going into the magnetic field. Thus:




vol

 
   dU EM 








E
J
dvol
E
H
dS


t 
S
where UEM is the electromagnetic field energy/volume. The left hand side is then the rate at
which energy is appearing in the volume V. Conservation of energy then requires that the right
hand side be the rate at which energy is entering the volume. Thus we can interpret:
  
S  EH

as the energy/sec/area carried by the electromagnetic field. S is called the “Poynting Vector”.
As an example of its use consider a current I flowing through a resistor R.
We know

Vin  Vout
EL
I
zˆ 
R
R
The current will produce a magnetic field:
  I
B  0 ˆ
2r
directed as shown. Then:

  B RI
 I
RI2
S  E

zˆ  0 ˆ  rˆ
0 L
2r0
2rL
Thus the energy entering the resistor/sec is:
S2rL  I 2R
as expected!