electromagnetic waves through a conducting media.The distance over which a plane wave is attenuated (reduce in amplitude) by a factor of 1/e is called skin depth.
E = π_0 π^(βπ·π).π^(π(πΌπ§βππ‘))
This represent an attenuated wave The attenuation is due to the Joule βloss equation(15) represent a plane wave traveling along z-direction .the imaginary part of k results in an attenuation of the wave.
3. Plane Electromagnetic waves in
conducting media
Maxwellβs equations are;
π». π· = π
π». π΅ = 0
π» Γ πΈ = β
ππ΅
ππ‘
π» Γ π» = π½ +
ππ·
ππ‘
Consider a conducting medium with permeability π
,permittivity π and conductivity π.
4. 1.Now Maxwell 1st
equation becomes
Charge density= π=0
π». π· = 0
As we know that
π· = ππΈ
π» . ππΈ =0
π π» . πΈ =0
Where π is not zero.
π». πΈ = 0
2. 2nd Maxwellβs
equation is
π». π΅ = 0
π΅ =ππ»
π» . ππ» =0
π π» . π» =0
Where π is not zero
π». π» = 0
3. 3rd Maxwellβs equation
becomes
π» Γ πΈ = β
ππ΅
ππ‘
where π΅
=ππ»
π» Γ πΈ = β
πππ»
ππ‘
π» Γ πΈ = βπ
ππ»
ππ‘
5. The 4th Maxwellβs equation is
π» Γ π» = π½ +
ππ·
ππ‘
where π½ = π πΈ and π· = ππΈ .so,
π» Γ π» = ππΈ +
πππΈ
ππ‘
π» Γ π» = ππΈ +π
ππΈ
ππ‘
So we get
π» Γ πΈ = βπ
ππ»
ππ‘
β¦β¦β¦β¦(1)
π» Γ π» = ππΈ +π
ππΈ
ππ‘
β¦β¦β¦.(2)
π». πΈ = 0 β¦β¦β¦β¦β¦β¦......(3)
π». π» = 0 β¦β¦β¦β¦β¦β¦β¦(4)
6.
7.
8. π»2
E β ππ
ππ
ππ‘
βππ
π2
π2
π
t
=0 β¦β¦..(7)
π»2π β ππ
ππ―
ππ‘
βππ
π2
π2
π―
t
=0 β¦β¦..(8)
Equation(7) and(8)are modified wave equations for
E
and H . So equation (7) is admits plane wave
solution
E = π0ππ(ππ₯βππ‘) _________-(9)
This can be easily checked by putting Eq.(9) in
Eq.(7)
-π2+ iπππ + π2ππ = 0 ------------------------(10)
Which show that wave numbers k must be complex
10. Equating real and imaginary parts, we get
πΌ2
β π½2
= π2
ππ __________(11)
2πΌπ½ β πππ = 0_________(12)
Solve the equation 12 for value of π½
2πΌπ½ =ΟΞΌΟ
π½ =
πππ
2πΌ
___________(13)
Put value of π½ ππ πππ’ππ‘πππ 11 π‘π πππ‘ π£πππ’π ππ πΌ
11. πΌ = π ππ[
1
2
Β±
1
2
1 +
π2
π2π2
1
2
]1/2
The negative sign would make πΌ complex so, we
Must choose the positive sign .so
πΌ = π ππ[
1
2
+
1
2
1 +
π2
π2π2
1
2
]1/2β¦β¦.(14)
Now, when k is positive equation (9) becomes
E = π0π ππΌ+π2π½ π§βπππ‘)
E = π0π (ππΌ+(β1)π½ π§βπππ‘)
E = π0π ππΌπ§βπ½π§ βπππ‘)
12. E = π0πβπ·π
.ππ(πΌπ§βππ‘)
___________(15)
This represent an attenuated wave The attenuation is
due to the Joule βloss equation(15) represent a plane
wave traveling along z-direction .the imaginary part
of k results in an attenuation of the wave.
Skin depth:
The distance over which a plane wave is attenuated
(reduce in amplitude) by a factor of 1/e is called skin
depth.
πΏ =
1
π½
13. It is measure to depth to which an electromagnetic
wave can penetrate the conductor.
Meanwhile, the real part of k determine the wave
length, the propagation speed and the index of
refraction in the usual way:
π =
2π
πΌ
π£ =
π
πΌ
π =
ππΌ
π
14. The quantity
ππΌ
π
is called complex index of
refraction
We call the material a poor conductor if
π
ππ
βͺ 1.
In this case
πΌ β ππ
πΌ β
π
2
π
π
And skin depth is independent of frequency
15. For a good conductor,
π
ππ
β« 1.
So we get
πΌ β π½ β
πππ
2
The skin depth decrease with increase in frequency .
The skin depth at optical frequencies (π β 1015π β1)
Is roughly 10β8π, which explain why metals are
opaque.