power n poynting theorem
TRANSCRIPT
POWER AND POYNTING THEOREM
Poynting Theorem
• It states that the “Net power flowing out of a given volume v is equal to the time rate of decrease in the energy stored within v minus the ohmic losses.”
• It mainly deals about the Energy-conservation(or Energy transportation).
Derivation of Poynting Vector
• The rate of such energy transportation can be obtained from Maxwell’s Equations.
#1 #2
t
t
• By doing the Dot Product on both sides of equation #1 with ‘E’,we get
#3• From the Vector Identity, • Applying this vector identity, #4
tE
tE
2
2
)()(
).()()(
)(
22
2
1Et
E
• By dotting the equation #1 with ‘H’,we get
• And thus the equation #4 becomes
• Rearranging terms and taking the volume integral on both sides,
t
EE
t
tt
22
2
2
1)(
2
)(2
)()(
dvEdvHEt
dvvvv
222
2
1
2
1)(
• Applying the Divergence theorem to the left-hand side of above equation gives
• This above Equation is referred to as Poynting’s Theorem.
vvS
dvEdvHEt
Sd 222
2
1
2
1)(
Total Power leavingThe Volume
Rate of decrease inEnergy stored in electric
and magnetic
Ohmic Powerdissipated-=
Poynting Vector
• The quantity “ExH” on the left-hand side of the previous equation is known as the POYNTING VECTOR P.
• It is measured in watts per square meter(W/m^2).
P =ExH
The Theorem is illustrated in this fig;
Illustration of power balance for EM fields
• The Poynting vector is normal to both ‘E’ and ‘H’ and is therefore along the direction of wave propagation ak.Thus,
• If we assume that
Then,
HEk aaa
yz
xz
azteE
tz
azteEtz
)cos(),(
)cos(),(
0
0
• So, by calculating P, we get
(1)
By integrating the Poynting vector P(z,t),over a period T=2π/w
(2)
zz aztzte
Etz )cos()cos(),( 2
20
P
zz azte
E)22cos(cos
22
20
T
avg dttzT
z0
),(1
)( PP
• By substituting the equation(1) in equation(2),we get
• The total time-average power crossing a given surface S is given by
zz
zavg aeE
cos
22
20
)(P
S
avgavg dSP P
PROBLEM• Question: A Uniform plane wave in a lossy
nonmagnetic medium has
(a) Compute the magnitude of the wave at z=4m,t=T/8. (b) Find the loss in dB suffered by the wave in the interval 0<z<3 m. (c) Find the Poynting vector at z=4,t=T/8.Take w=10^8 rad/sec.
mjeaa zyxs /4.32.0,)125(
Solution (a)
This Formula was derived separately by the concept of Wave Propagation in Lossy Dielectrics…
At z=4 and t=T/8,w t = (2π/T)x(T/8) = π/4
)Re()(Re),( )(0 x
ztjztjs aeeEezEtz
)cos(),( 0 zteEtz z
)4.3cos(125Re 2.0 zteaaeE zyx
jwts
6.134
cos)125( 8.0 eaa yx
662.5)6.134/cos(13 8.0 e
• (b) Loss = α∆z = 0.2(3) = 0.6 Np Loss in dB = 0.6x8.686 = 5.212 dB (1
Np=8.686dB)
. (c) To find the Poynting vector,P
P = ExH where η=Intrinsic Impedence
x
kx a
• The formula for Intrinsic Impedance ‘η’ is given by
let = x x=1.00694
4/12
1
/
2
1
17/14.3/2.02
1x1-x
5.3200694.152.11
120
1
0
0
x
r
118.012tan 2x
0365.3
0365.35.32
zjyxzyx
zskx ee
aaeaa
aa
0365.3125
125
mAzteaa zyx )365.34.3cos(8.1532.369 02.0
)365.34.3cos()4.3cos(1008.1532.369
0125 04.03
ztzte z P
)365.34.3cos()4.3cos(2.5 04.0 ztzte z PBy Substituting the values of z=4 and t=T/8,
2/9702.0 mWazPPoynting Vector,