Obtain the Poynting theorem for conservation of energy in an electromagnetic field?

Answer 1

Poynting's Theorem

In previous classes we found that the work necessary to assemble a static

charge distribution (against the Coulomb repulsion of like charges) is

ε

W 0 E2dτ e 2

where E is the resulting electric field.

Likewise, the work required to get currents going (against the back emf) is

1 2 Wm B dτ 2μ

0

where B is the resulting magnetic field.

This suggests that the total energy stored in electromagnetic fields is

1 2 1 2 Uem ε E B dτ 2 0 μ

0

------------ (5)

I propose to derive Equation 5 more generally, now, in the context of the

energy conservation law for electrodynamics.

Suppose we have some charge and current configuration which, at time t,

produces fields E and B. In the next instant, dt, the charges move around a

bit. Question: How much work, dW, is done by the electromagnetic forces

acting on these charges in the interval dt? According to the Lorentz force

law, the work done on a charge q is

F. dI = q (E + v x B). v dt = qE . v dt.

Now, q = ρdτ and ρv = J, so the rate at which work is done on all the

charges in a volume V is

dW

dτ

dt V

E J ------------- (6)

Evidently E Jis the work done per unit time, per unit volume-which is to

say, the power delivered per unit volume. We can express this quantity in

terms of the fields alone, using the Ampere-Maxwell law to eliminate J:

1

( ) ε

μ 0

0

t

E

E J E B E

From product rule 6,

E B B ( E) E ( B)

Invoking Faraday‟s law( E B/ t) , it follows that

( ) ( )

t

B

E B B E B

Mean while,

1 2 (B )

t 2 t

B

B and

1 2 (E )

t 2 t

E

E -------------- (7)

so

1 2 1 2 1 ε E B ( )

2 0 μ μ

0 0

t

E J E B ------------- (8)

Putting this into Equation 6, and applying the divergence theorem to the

second term, we have

dW d 1 2 1 2 1 ε E B dτ ( ) d

dt dt 2 0 μ μ V 0 0 S

 E B a ---------- (9)

where S is the surface bounding V. This is Poynting's theorem;