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Electricity and Magnetism-2
Electromagnetic waves in
conducting media
Plane Electromagnetic waves in
conducting media
Maxwell’s equations are;
𝛻. 𝐷 = 𝜌
𝛻. 𝐡 = 0
𝛻 Γ— 𝐸 = βˆ’
πœ•π΅
πœ•π‘‘
𝛻 Γ— 𝐻 = 𝐽 +
πœ•π·
πœ•π‘‘
Consider a conducting medium with permeability πœ‡
,permittivity πœ€ and conductivity 𝜎.
1.Now Maxwell 1st
equation becomes
Charge density= 𝜌=0
𝛻. 𝐷 = 0
As we know that
𝐷 = πœ€πΈ
𝛻 . πœ€πΈ =0
πœ€ 𝛻 . 𝐸 =0
Where πœ€ is not zero.
𝛻. 𝐸 = 0
2. 2nd Maxwell’s
equation is
𝛻. 𝐡 = 0
𝐡 =πœ‡π»
𝛻 . πœ‡π» =0
πœ‡ 𝛻 . 𝐻 =0
Where πœ‡ is not zero
𝛻. 𝐻 = 0
3. 3rd Maxwell’s equation
becomes
𝛻 Γ— 𝐸 = βˆ’
πœ•π΅
πœ•π‘‘
where 𝐡
=πœ‡π»
𝛻 Γ— 𝐸 = βˆ’
πœ•πœ‡π»
πœ•π‘‘
𝛻 Γ— 𝐸 = βˆ’πœ‡
πœ•π»
πœ•π‘‘
The 4th Maxwell’s equation is
𝛻 Γ— 𝐻 = 𝐽 +
πœ•π·
πœ•π‘‘
where 𝐽 = 𝜎 𝐸 and 𝐷 = πœ€πΈ .so,
𝛻 Γ— 𝐻 = 𝜎𝐸 +
πœ•πœ€πΈ
πœ•π‘‘
𝛻 Γ— 𝐻 = 𝜎𝐸 +πœ€
πœ•πΈ
πœ•π‘‘
So we get
𝛻 Γ— 𝐸 = βˆ’πœ‡
πœ•π»
πœ•π‘‘
…………(1)
𝛻 Γ— 𝐻 = 𝜎𝐸 +πœ€
πœ•πΈ
πœ•π‘‘
……….(2)
𝛻. 𝐸 = 0 ………………......(3)
𝛻. 𝐻 = 0 …………………(4)
𝛻2
E βˆ’ πœ‡πœŽ
πœ•π„
πœ•π‘‘
βˆ’πœ‡πœ–
πœ•2
πœ•2
𝐄
t
=0 ……..(7)
𝛻2𝐇 βˆ’ πœ‡πœŽ
πœ•π‘―
πœ•π‘‘
βˆ’πœ‡πœ–
πœ•2
πœ•2
𝑯
t
=0 ……..(8)
Equation(7) and(8)are modified wave equations for
E
and H . So equation (7) is admits plane wave
solution
E = 𝐄0𝑒𝑖(π‘˜π‘₯βˆ’πœ”π‘‘) _________-(9)
This can be easily checked by putting Eq.(9) in
Eq.(7)
-π‘˜2+ iπœ”πœ‡πœŽ + πœ”2πœ–πœ‡ = 0 ------------------------(10)
Which show that wave numbers k must be complex
Let k = 𝛼 +iΞ².then from equation (10) we get
βˆ’[𝛼+i Ξ²] 2
+ iπœ”πœ‡πœŽ + πœ”2πœ–πœ‡ = 0
βˆ’ 𝛼2 + βˆ’1 𝛽2 + 2iα𝛽 + iπœ”πœ‡πœŽ + πœ”2πœ–πœ‡ = 0
βˆ’π›Ό2 + 𝛽2 βˆ’ 2iα𝛽 + iπœ”πœ‡πœŽ + πœ”2πœ–πœ‡ = 0
𝛼2 + 𝛽2 +(2𝛼𝛽 βˆ’ πœ”πœ‡πœŽ)𝑖 βˆ’ πœ”2πœ–πœ‡ = 0
𝛼2
βˆ’ 𝛽2
+(2𝛼𝛽 βˆ’ πœ”πœ‡πœŽ)𝑖 = πœ”2
πœ–πœ‡ + 0𝑖
Equating real and imaginary parts, we get
𝛼2
βˆ’ 𝛽2
= πœ”2
πœ–πœ‡ __________(11)
2𝛼𝛽 βˆ’ πœ”πœ‡πœŽ = 0_________(12)
Solve the equation 12 for value of 𝛽
2𝛼𝛽 =ωμσ
𝛽 =
πœ”πœ‡πœŽ
2𝛼
___________(13)
Put value of 𝛽 𝑖𝑛 π‘’π‘žπ‘’π‘Žπ‘‘π‘–π‘œπ‘› 11 π‘‘π‘œ 𝑔𝑒𝑑 π‘£π‘Žπ‘™π‘’π‘’ π‘œπ‘“ 𝛼
𝛼 = πœ” πœ€πœ‡[
1
2
Β±
1
2
1 +
𝜎2
πœ”2πœ€2
1
2
]1/2
The negative sign would make 𝛼 complex so, we
Must choose the positive sign .so
𝛼 = πœ” πœ€πœ‡[
1
2
+
1
2
1 +
𝜎2
πœ”2πœ€2
1
2
]1/2…….(14)
Now, when k is positive equation (9) becomes
E = 𝐄0𝑒 𝑖𝛼+𝑖2𝛽 π‘§βˆ’π‘–πœ”π‘‘)
E = 𝐄0𝑒 (𝑖𝛼+(βˆ’1)𝛽 π‘§βˆ’π‘–πœ”π‘‘)
E = 𝐄0𝑒 π‘–π›Όπ‘§βˆ’π›½π‘§ βˆ’π‘–πœ”π‘‘)
E = 𝐄0π’†βˆ’πœ·π’›
.𝑒𝑖(π›Όπ‘§βˆ’πœ”π‘‘)
___________(15)
This represent an attenuated wave The attenuation is
due to the Joule –loss 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.
𝛿 =
1
𝛽
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œ† =
2πœ‹
𝛼
𝑣 =
πœ”
𝛼
𝑛 =
𝑐𝛼
πœ”
The quantity
𝑐𝛼
πœ”
is called complex index of
refraction
We call the material a poor conductor if
𝜎
πœ”πœ€
β‰ͺ 1.
In this case
𝛼 β‰ˆ πœ€πœ‡
𝛼 β‰ˆ
𝜎
2
πœ‡
πœ”
And skin depth is independent of frequency
For a good conductor,
𝜎
πœ”πœ€
≫ 1.
So we get
𝛼 β‰ˆ 𝛽 β‰ˆ
πœ”πœ‡πœŽ
2
The skin depth decrease with increase in frequency .
The skin depth at optical frequencies (πœ” β‰ˆ 1015π‘ βˆ’1)
Is roughly 10βˆ’8π‘š, which explain why metals are
opaque.
E.M WAVES IN CONDUCTOR MEDIA.pptx

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E.M WAVES IN CONDUCTOR MEDIA.pptx

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  • 3. Plane Electromagnetic waves in conducting media Maxwell’s equations are; 𝛻. 𝐷 = 𝜌 𝛻. 𝐡 = 0 𝛻 Γ— 𝐸 = βˆ’ πœ•π΅ πœ•π‘‘ 𝛻 Γ— 𝐻 = 𝐽 + πœ•π· πœ•π‘‘ Consider a conducting medium with permeability πœ‡ ,permittivity πœ€ and conductivity 𝜎.
  • 4. 1.Now Maxwell 1st equation becomes Charge density= 𝜌=0 𝛻. 𝐷 = 0 As we know that 𝐷 = πœ€πΈ 𝛻 . πœ€πΈ =0 πœ€ 𝛻 . 𝐸 =0 Where πœ€ is not zero. 𝛻. 𝐸 = 0 2. 2nd Maxwell’s equation is 𝛻. 𝐡 = 0 𝐡 =πœ‡π» 𝛻 . πœ‡π» =0 πœ‡ 𝛻 . 𝐻 =0 Where πœ‡ is not zero 𝛻. 𝐻 = 0 3. 3rd Maxwell’s equation becomes 𝛻 Γ— 𝐸 = βˆ’ πœ•π΅ πœ•π‘‘ where 𝐡 =πœ‡π» 𝛻 Γ— 𝐸 = βˆ’ πœ•πœ‡π» πœ•π‘‘ 𝛻 Γ— 𝐸 = βˆ’πœ‡ πœ•π» πœ•π‘‘
  • 5. The 4th Maxwell’s equation is 𝛻 Γ— 𝐻 = 𝐽 + πœ•π· πœ•π‘‘ where 𝐽 = 𝜎 𝐸 and 𝐷 = πœ€πΈ .so, 𝛻 Γ— 𝐻 = 𝜎𝐸 + πœ•πœ€πΈ πœ•π‘‘ 𝛻 Γ— 𝐻 = 𝜎𝐸 +πœ€ πœ•πΈ πœ•π‘‘ So we get 𝛻 Γ— 𝐸 = βˆ’πœ‡ πœ•π» πœ•π‘‘ …………(1) 𝛻 Γ— 𝐻 = 𝜎𝐸 +πœ€ πœ•πΈ πœ•π‘‘ ……….(2) 𝛻. 𝐸 = 0 ………………......(3) 𝛻. 𝐻 = 0 …………………(4)
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  • 8. 𝛻2 E βˆ’ πœ‡πœŽ πœ•π„ πœ•π‘‘ βˆ’πœ‡πœ– πœ•2 πœ•2 𝐄 t =0 ……..(7) 𝛻2𝐇 βˆ’ πœ‡πœŽ πœ•π‘― πœ•π‘‘ βˆ’πœ‡πœ– πœ•2 πœ•2 𝑯 t =0 ……..(8) Equation(7) and(8)are modified wave equations for E and H . So equation (7) is admits plane wave solution E = 𝐄0𝑒𝑖(π‘˜π‘₯βˆ’πœ”π‘‘) _________-(9) This can be easily checked by putting Eq.(9) in Eq.(7) -π‘˜2+ iπœ”πœ‡πœŽ + πœ”2πœ–πœ‡ = 0 ------------------------(10) Which show that wave numbers k must be complex
  • 9. Let k = 𝛼 +iΞ².then from equation (10) we get βˆ’[𝛼+i Ξ²] 2 + iπœ”πœ‡πœŽ + πœ”2πœ–πœ‡ = 0 βˆ’ 𝛼2 + βˆ’1 𝛽2 + 2iα𝛽 + iπœ”πœ‡πœŽ + πœ”2πœ–πœ‡ = 0 βˆ’π›Ό2 + 𝛽2 βˆ’ 2iα𝛽 + iπœ”πœ‡πœŽ + πœ”2πœ–πœ‡ = 0 𝛼2 + 𝛽2 +(2𝛼𝛽 βˆ’ πœ”πœ‡πœŽ)𝑖 βˆ’ πœ”2πœ–πœ‡ = 0 𝛼2 βˆ’ 𝛽2 +(2𝛼𝛽 βˆ’ πœ”πœ‡πœŽ)𝑖 = πœ”2 πœ–πœ‡ + 0𝑖
  • 10. Equating real and imaginary parts, we get 𝛼2 βˆ’ 𝛽2 = πœ”2 πœ–πœ‡ __________(11) 2𝛼𝛽 βˆ’ πœ”πœ‡πœŽ = 0_________(12) Solve the equation 12 for value of 𝛽 2𝛼𝛽 =ωμσ 𝛽 = πœ”πœ‡πœŽ 2𝛼 ___________(13) Put value of 𝛽 𝑖𝑛 π‘’π‘žπ‘’π‘Žπ‘‘π‘–π‘œπ‘› 11 π‘‘π‘œ 𝑔𝑒𝑑 π‘£π‘Žπ‘™π‘’π‘’ π‘œπ‘“ 𝛼
  • 11. 𝛼 = πœ” πœ€πœ‡[ 1 2 Β± 1 2 1 + 𝜎2 πœ”2πœ€2 1 2 ]1/2 The negative sign would make 𝛼 complex so, we Must choose the positive sign .so 𝛼 = πœ” πœ€πœ‡[ 1 2 + 1 2 1 + 𝜎2 πœ”2πœ€2 1 2 ]1/2…….(14) Now, when k is positive equation (9) becomes E = 𝐄0𝑒 𝑖𝛼+𝑖2𝛽 π‘§βˆ’π‘–πœ”π‘‘) E = 𝐄0𝑒 (𝑖𝛼+(βˆ’1)𝛽 π‘§βˆ’π‘–πœ”π‘‘) E = 𝐄0𝑒 π‘–π›Όπ‘§βˆ’π›½π‘§ βˆ’π‘–πœ”π‘‘)
  • 12. E = 𝐄0π’†βˆ’πœ·π’› .𝑒𝑖(π›Όπ‘§βˆ’πœ”π‘‘) ___________(15) This represent an attenuated wave The attenuation is due to the Joule –loss 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. 𝛿 = 1 𝛽
  • 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œ† = 2πœ‹ 𝛼 𝑣 = πœ” 𝛼 𝑛 = 𝑐𝛼 πœ”
  • 14. The quantity 𝑐𝛼 πœ” is called complex index of refraction We call the material a poor conductor if 𝜎 πœ”πœ€ β‰ͺ 1. In this case 𝛼 β‰ˆ πœ€πœ‡ 𝛼 β‰ˆ 𝜎 2 πœ‡ πœ” And skin depth is independent of frequency
  • 15. For a good conductor, 𝜎 πœ”πœ€ ≫ 1. So we get 𝛼 β‰ˆ 𝛽 β‰ˆ πœ”πœ‡πœŽ 2 The skin depth decrease with increase in frequency . The skin depth at optical frequencies (πœ” β‰ˆ 1015π‘ βˆ’1) Is roughly 10βˆ’8π‘š, which explain why metals are opaque.