Page 402 - Engineering Electromagnetics, 8th Edition

384 ENGINEERING ELECTROMAGNETICS

D11.4. Given a nonmagnetic material having = 3.2 and σ = 1.5 × 10 −4

r
S/m, find numerical values at 3 MHz for the (a) loss tangent; (b) attenuation
constant; (c) phase constant; (d) intrinsic impedance.
◦
Ans. 0.28; 0.016 Np/m; 0.11 rad/m; 207 7.8
D11.5. Consider a material for which µ r = 1, = 2.5, and the loss tangent

r
is 0.12. If these three values are constant with frequency in the range 0.5 MHz ≤
f ≤ 100 MHz, calculate: (a) σ at 1 and 75 MHz; (b) λ at 1 and 75 MHz; (c) ν p
at 1 and 75 MHz.
8
Ans. 1.67 × 10 −5 and 1.25 × 10 −3 S/m; 190 and 2.53 m; 1.90 × 10 m/s twice

11.3 POYNTING’S THEOREM
AND WAVE POWER
In order to find the power flow associated with an electromagnetic wave, it is necessary
to develop a power theorem for the electromagnetic field known as the Poynting the-
orem. It was originally postulated in 1884 by an English physicist, John H. Poynting.
The development begins with one of Maxwell’s curl equations, in which we
assume that the medium may be conductive:
∂D
∇× H = J + (63)
∂t
Next, we take the scalar product of both sides of (63) with E,
∂D
E · ∇× H = E · J + E · (64)
∂t
We then introduce the following vector identity, which may be proved by expansion
in rectangular coordinates:
∇ · (E × H) =−E · ∇× H + H · ∇× E (65)
Using (65) in the left side of (64) results in
∂D
H · ∇× E −∇ · (E × H) = J · E + E · (66)
∂t
where the curl of the electric field is given by the other Maxwell curl equation:
∂B
∇× E =−
∂t
Therefore
∂B ∂D
−H · −∇ · (E × H) = J · E + E ·
∂t ∂t
or
∂E ∂H
−∇ · (E × H) = J · E + E · + µH · (67)
∂t ∂t