Electromagnetic Waves

Lecture 35: Electromagnetic Theory

Professor D. K. Ghosh, Physics Department, I.I.T., Bombay

Propagation of Electromagnetic Waves in a Conducting Medium We will consider a plane electromagnetic wave travelling in a linear dielectric medium such as air along the z direction and being incident at a conducting interface. The medium will be taken to be a linear medium. So that one can describe the electrodynamics using only the E and H vectors. We wish to investigate the propagation of the wave in the conducting medium.

As the medium is linear and the propagation takes place in the infinite medium, the vectors , and are still mutually perpendicular. We take the electric field along the x direction, the magnetic field along 𝐸𝐸�⃗ 𝐻𝐻��⃗ 𝑘𝑘�⃗ the y- dirrection and the propagation to take place in the z direction. Further, we will take the conductivity to be finite and the conductor to obey Ohm’s law, = . Consider the pair of curl equations of Maxwell. 𝐽𝐽⃗ 𝜎𝜎𝐸𝐸�⃗

× = = 𝜕𝜕𝐵𝐵�⃗ 𝜕𝜕𝐻𝐻��⃗ 𝛻𝛻 𝐸𝐸�⃗ − −μ × = +𝜕𝜕𝜕𝜕 = 𝜕𝜕𝜕𝜕 + 𝜕𝜕𝐸𝐸�⃗ 𝜕𝜕𝐸𝐸�⃗ 𝛻𝛻 𝐻𝐻��⃗ 𝐽𝐽⃗ 𝜖𝜖 𝜎𝜎𝐸𝐸�⃗ 𝜖𝜖 Let us take , and to be respectively in x, y and z𝜕𝜕𝜕𝜕 direction. We𝜕𝜕𝜕𝜕 the nhave,

𝐸𝐸�⃗ 𝐻𝐻��⃗ 𝑘𝑘�⃗ × = = 𝜕𝜕𝐸𝐸𝑥𝑥 𝜕𝜕𝐻𝐻𝑦𝑦 �𝛻𝛻 𝐸𝐸�⃗�𝑦𝑦 −μ i.e., 𝜕𝜕𝜕𝜕 𝜕𝜕𝜕𝜕

+ = 0 (1) 𝜕𝜕𝐸𝐸𝑥𝑥 𝜕𝜕𝐻𝐻𝑦𝑦 μ and 𝜕𝜕𝜕𝜕 𝜕𝜕𝜕𝜕

× = = + 𝜕𝜕𝐻𝐻𝑦𝑦 𝜕𝜕𝐸𝐸𝑥𝑥 �𝛻𝛻 𝐻𝐻��⃗�𝑥𝑥 − 𝜎𝜎𝐸𝐸𝑥𝑥 𝜖𝜖 i.e. 𝜕𝜕𝜕𝜕 𝜕𝜕𝜕𝜕

+ + = 0 (2) 𝜕𝜕𝐻𝐻𝑦𝑦 𝜕𝜕𝐸𝐸𝑥𝑥 𝜎𝜎𝐸𝐸𝑥𝑥 𝜖𝜖 𝜕𝜕𝜕𝜕 𝜕𝜕𝜕𝜕

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We take the time variation to be harmonic ( ~ ) so that the time derivative is equivalent to a multiplication by . The pair of equations (1) and𝑖𝑖𝑖𝑖𝑖𝑖 (2) can then be written as 𝑒𝑒

𝑖𝑖𝑖𝑖 + Hy = 0 𝜕𝜕𝐸𝐸𝑥𝑥 𝑖𝑖μω 𝜕𝜕𝜕𝜕 + + = 0 𝜕𝜕𝐻𝐻𝑦𝑦 We can solve this pair of coupled equations𝜎𝜎 by𝐸𝐸𝑥𝑥 taking𝑖𝑖𝑖𝑖𝑖𝑖 a𝐸𝐸 𝑥𝑥derivative of either of the equations with respect to z and substituting the other𝜕𝜕𝜕𝜕 into it, 2 2 Hy 2 + = 2 ( + ) = 0 𝜕𝜕 𝐸𝐸𝑥𝑥 ∂ 𝜕𝜕 𝐸𝐸𝑥𝑥 𝑖𝑖μω − 𝑖𝑖𝑖𝑖𝑖𝑖 𝜎𝜎 𝑖𝑖𝑖𝑖𝑖𝑖 𝐸𝐸𝑥𝑥 Define, a complex constant𝜕𝜕𝑧𝑧 γ through𝜕𝜕𝜕𝜕 𝜕𝜕𝑧𝑧 2 = ( + ) in terms of which we have, 2𝛾𝛾 𝑖𝑖𝑖𝑖𝑖𝑖 𝜎𝜎 𝑖𝑖𝑖𝑖𝑖𝑖 2 2 = 0 (3) 𝜕𝜕 𝐸𝐸𝑥𝑥 In an identical fashion, we get − 𝛾𝛾 𝐸𝐸𝑥𝑥 𝜕𝜕𝑧𝑧 2 2 2 = 0 (4) 𝜕𝜕 𝐻𝐻𝑦𝑦 − 𝛾𝛾 𝐻𝐻𝑦𝑦 Solutions of (3) and (4) are well known𝜕𝜕𝑧𝑧 and are expressed in terms of hyperbolic functions, = cosh( ) + sinh( ) = cosh( ) + sinh( ) 𝐸𝐸𝑥𝑥 𝐴𝐴 𝛾𝛾𝛾𝛾 𝐵𝐵 𝛾𝛾𝛾𝛾 where A, B, C and D are constants to be determined. If the values of the electric field at z=0 is 𝐻𝐻𝑦𝑦 𝐶𝐶 𝛾𝛾𝛾𝛾 𝐷𝐷 𝛾𝛾𝛾𝛾 0 and that of the magnetic field at z=0 is 0, we have = 0 and = 0. In order to determine the constants B and D, let us return back to the original first order 𝐸𝐸 𝐻𝐻 𝐴𝐴 𝐸𝐸 𝐶𝐶 𝐻𝐻 equations (1) and (2)

+ Hy = 0 𝜕𝜕𝐸𝐸𝑥𝑥 𝑖𝑖μω 𝜕𝜕𝜕𝜕 + + = 0 𝜕𝜕𝐻𝐻𝑦𝑦 Substituting the solutions for E and H 𝜎𝜎𝐸𝐸𝑥𝑥 𝑖𝑖𝑖𝑖𝑖𝑖𝐸𝐸𝑥𝑥 𝜕𝜕𝜕𝜕 0 sinh( ) + cosh( ) + (H0cosh( ) + sinh( )) = 0 This equation must remain valid for all values of z, which is possible if the coefficients of 𝛾𝛾𝐸𝐸 𝛾𝛾𝛾𝛾 𝐵𝐵𝐵𝐵 𝛾𝛾𝛾𝛾 𝑖𝑖𝑖𝑖𝑖𝑖 𝛾𝛾𝛾𝛾 𝐷𝐷 𝛾𝛾𝛾𝛾 and cosh terms are separately equated to zero, 𝑠𝑠𝑠𝑠𝑠𝑠ℎ 0 + = 0 + 0 = 0 𝐸𝐸 𝛾𝛾 𝑖𝑖𝑖𝑖𝑖𝑖𝑖𝑖 The former gives, 𝐵𝐵𝐵𝐵 𝑖𝑖𝑖𝑖𝑖𝑖𝐻𝐻

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= 0 𝛾𝛾 𝐷𝐷 − (𝐸𝐸 + ) = 𝑖𝑖𝑖𝑖𝑖𝑖 �𝑖𝑖𝑖𝑖𝑖𝑖 𝜎𝜎 𝑖𝑖𝑖𝑖𝑖𝑖 − + 𝑖𝑖𝑖𝑖𝑖𝑖 = 0 𝜎𝜎 𝑖𝑖𝑖𝑖𝑖𝑖 −�0 𝐸𝐸 = 𝑖𝑖𝑖𝑖𝑖𝑖 𝐸𝐸 where − 𝜂𝜂 = + 𝑖𝑖𝑖𝑖𝑖𝑖 𝜂𝜂 � 𝜎𝜎 𝑖𝑖𝑖𝑖𝑖𝑖

Likewise, we get,

= 0 Substituting these, our solutions for the E and H become, 𝐵𝐵 −𝜂𝜂𝐻𝐻

= 0 cosh( ) 0 sinh( ) 0 𝐸𝐸𝑥𝑥 = 𝐸𝐸 0 cosh(𝛾𝛾𝛾𝛾 )− 𝜂𝜂𝐻𝐻 sinh( 𝛾𝛾𝛾𝛾) 𝐸𝐸 The wave is propagating in the 𝐻𝐻z 𝑦𝑦direction.𝐻𝐻 Let 𝛾𝛾us𝛾𝛾 evaluate− the𝛾𝛾 fields𝛾𝛾 when the wave has 𝜂𝜂 reached = ,

= 0 cosh( ) 0 sinh( ) 𝑧𝑧 𝑙𝑙 0 𝐸𝐸𝑥𝑥 = 𝐸𝐸 0 cosh(𝛾𝛾𝛾𝛾 )− 𝜂𝜂𝐻𝐻 sinh( 𝛾𝛾)𝛾𝛾 𝐸𝐸 If is large, we can approximate𝐻𝐻 𝑦𝑦 𝐻𝐻 𝛾𝛾𝛾𝛾 − 𝛾𝛾𝛾𝛾 𝜂𝜂 𝑙𝑙 sinh( ) cosh( ) = 2𝛾𝛾𝛾𝛾 𝑒𝑒 we then have, 𝛾𝛾𝛾𝛾 ≈ 𝛾𝛾𝛾𝛾 = ( ) 0 0 2𝛾𝛾𝛾𝛾 𝑒𝑒 𝑥𝑥 0 𝐸𝐸 = (𝐸𝐸 − 𝜂𝜂𝐻𝐻) 0 2𝛾𝛾𝛾𝛾 𝐸𝐸 𝑒𝑒 The ratio of the magnitudes of the electric𝐻𝐻𝑦𝑦 field𝐻𝐻 to− magnetic field is defined as the “characteristic 𝜂𝜂 impedance” of the wave

= = + 𝐸𝐸𝑥𝑥 𝑖𝑖𝑖𝑖𝑖𝑖 � � 𝜂𝜂 � 𝐻𝐻𝑦𝑦 𝜎𝜎 𝑖𝑖𝑖𝑖𝑖𝑖 Suppose we have lossless medium, σ=0, i.e. for a perfect conductor, the characteristic impedance is

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= 7 𝜇𝜇 2 12 2 2 If the medium is vacuum, = 0 = 4 × 10𝜂𝜂 N�A and = 0 = 8.85 × 10 C N m 𝜖𝜖 gives 377 . The characteristic impedance,− as the name suggests, has the −dimension of 𝜇𝜇 𝜇𝜇 𝜋𝜋 ⁄ 𝜖𝜖 𝜖𝜖 ⁄ − resistance. 𝜂𝜂 ≈ 𝛺𝛺 In this case, = .

Let us look at the full𝛾𝛾 three𝑖𝑖√ 𝜇𝜇𝜇𝜇dimensional version of the propagation in a conductor. Once again, we start with the two curl equations,

× = 𝜕𝜕𝐻𝐻��⃗ 𝛻𝛻 𝐸𝐸�⃗ −μ × = +𝜕𝜕𝜕𝜕 𝜕𝜕𝐸𝐸�⃗ 𝛻𝛻 𝐻𝐻��⃗ 𝜎𝜎𝐸𝐸�⃗ 𝜖𝜖 Take a curl of both sides of the first equation, 𝜕𝜕𝜕𝜕

( × ) × × = 2 = 𝜕𝜕 𝛻𝛻 𝐻𝐻��⃗ 𝛻𝛻 �𝛻𝛻 𝐸𝐸�⃗� 𝛻𝛻�𝛻𝛻 ⋅ 𝐸𝐸�⃗� − 𝛻𝛻 𝐸𝐸�⃗ −μ As there are no charges or currents, we ignore the divergence term and𝜕𝜕𝜕𝜕 substitute for the curl of H from the second equation,

2 = + 𝜕𝜕 𝜕𝜕𝐸𝐸�⃗ 𝛻𝛻 𝐸𝐸�⃗ μ �𝜎𝜎𝐸𝐸�⃗ 𝜖𝜖 � 𝜕𝜕𝜕𝜕 2 𝜕𝜕𝜕𝜕 = + 2 𝜕𝜕𝐸𝐸�⃗ 𝜕𝜕 𝐸𝐸�⃗ 𝜎𝜎𝜎𝜎 𝜎𝜎𝜎𝜎 We take the propagating solutions to be 𝜕𝜕𝜕𝜕 𝜕𝜕𝑡𝑡

= 0 𝑖𝑖�𝑘𝑘�⃗⋅𝑟𝑟⃗−𝜔𝜔𝜔𝜔 � so that the above equation becomes, 𝐸𝐸�⃗ 𝐸𝐸�⃗ 𝑒𝑒

2 = ( + 2 ) so that we have, the complex propagation𝑘𝑘 𝐸𝐸constant�⃗ 𝑖𝑖𝑖𝑖𝑖𝑖 to𝑖𝑖 be𝜔𝜔 given𝜇𝜇𝜇𝜇 𝐸𝐸�by⃗

2 = + 2 so that 𝑘𝑘 𝑖𝑖𝑖𝑖𝑖𝑖𝑖𝑖 𝜔𝜔 𝜇𝜇𝜇𝜇

= ( + ) k is complex and its real and imaginary parts𝑘𝑘 can� 𝜔𝜔𝜔𝜔be separ𝜔𝜔𝜔𝜔 ated𝑖𝑖𝑖𝑖 by standard algebra,

4 we have

1/2 1/2 2 2 = 1 + 1 + + 1 + 1 2 2 2 2 2 𝜇𝜇𝜇𝜇 𝜎𝜎 𝜎𝜎 𝑘𝑘 𝜔𝜔� �� � � 𝑖𝑖 �� − � � 𝜔𝜔 𝜖𝜖 𝜔𝜔 𝜖𝜖 Thus the propagation vector β and the attenuation factor α are given by

2 = 1 + + 1 2 2 2 𝜇𝜇𝜇𝜇 𝜎𝜎 𝛽𝛽 𝜔𝜔� ��� � 𝜔𝜔 𝜖𝜖 2 = 1 + 1 2 2 2 𝜇𝜇𝜇𝜇 𝜎𝜎 𝛼𝛼 𝜔𝜔� ��� − � 𝜔𝜔 𝜖𝜖 The ration determines whether a material is a good conductor or otherwise. Consider a good 𝜎𝜎 conductor for𝜔𝜔𝜔𝜔 which . For this case, we have, 𝜎𝜎 ≫ 𝜔𝜔𝜔𝜔 = = 2 𝜔𝜔𝜔𝜔𝜔𝜔 𝛽𝛽 𝛼𝛼 � The speed of electromagnetic wave is given by

2 = = 𝜔𝜔 𝜔𝜔 𝑣𝑣 � 𝛽𝛽 𝜎𝜎𝜎𝜎 The electric field amplitude diminishes with distance as . The distance to which the field penetrates before its amplitude diminishes be a factor 1 is known −as𝛼𝛼 𝛼𝛼the “skin depth” , which is given by 𝑒𝑒 − 𝑒𝑒 1 2 = =

𝛿𝛿 � 𝛼𝛼 𝜔𝜔𝜔𝜔𝜔𝜔 The wave does not penetrate much inside a conductor. Consider electromagnetic wave of frequency 1 MHz for copper which has a conductivity of approximately 6 × 107 1m 1. Substituting these values, one gets the skin depth in Cu to be about 0.067 mm. For comparison,− the− skin depth in sea water which 𝛺𝛺 is conducting because of salinity, is about 25 cm while that for fresh water is nearly 7m. Because of small skin depth in conductors, any current that arises in the metal because of the electromagnetic wave is confined within a thin layer of the surface.

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Reflection and Transmission from interface of a conductor

Consider an electromagnetic wave to be incident normally at the interface between a dielectric and a conductor. As before, we take the media to be linear and assume no charge or current densities to exist anywhere. We then have a continuity of the electric and the magnetic fields at the interface so that

+ = + = 𝐸𝐸𝐼𝐼 𝐸𝐸𝑅𝑅 𝐸𝐸𝑇𝑇 The relationships between the magnetic field𝐻𝐻 and𝐼𝐼 the𝐻𝐻𝑅𝑅 electric𝐻𝐻𝑇𝑇 field are given by

= 1 𝐸𝐸𝐼𝐼 𝐻𝐻𝐼𝐼 =𝜂𝜂 1 𝐸𝐸𝑅𝑅 𝐻𝐻𝑅𝑅 − = 𝜂𝜂 2 𝐸𝐸𝑇𝑇 𝐻𝐻𝑇𝑇 the minus sign in the second relation comes because 𝜂𝜂of the propagation direction having been reversed on reflection.

Solving these, we get,

= 2 1 2 + 1 𝐸𝐸𝑅𝑅 𝜂𝜂 − 𝜂𝜂 2 2 𝐸𝐸𝐼𝐼 = 𝜂𝜂 =𝜂𝜂 2 + 1 𝐸𝐸𝑇𝑇 𝐸𝐸𝑅𝑅 𝜂𝜂 𝐼𝐼 𝐼𝐼 The magnetic field expressions are given by𝐸𝐸 interchanging𝐸𝐸 𝜂𝜂 1𝜂𝜂 and 2 in the above expressions.

𝜂𝜂 𝜂𝜂

E

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Let us look at consequence of this. Consider a good conductor such as copper. We can see that 2 is a small complex number. For instance, taking the wave frequency to be 1 MHz and substituting 6 1 1 𝜂𝜂 conductivity of Cu to be 6 × 10 , we can calculate 2 to be approximately (1 + ) × 2.57 × 4 10 whereas the vacuum impedance− − 1 = 377 . This implies 𝛺𝛺 𝑚𝑚 𝜂𝜂 𝑖𝑖 − 𝛺𝛺 𝜂𝜂 2𝛺𝛺 1 = 1 2 + 1 𝐸𝐸𝑅𝑅 𝜂𝜂 − 𝜂𝜂 ≈ − which shows that a good metal is also a good𝐸𝐸𝐼𝐼 reflector.𝜂𝜂 𝜂𝜂 On the other hand, if we calculate the transmission coefficient we find it to be substantially reduced, being only about (1 + ) × 1 6. − For the transmitted magnetic field, the ratio is approximately +2. Though E is reflected𝑖𝑖 with a change 𝐻𝐻𝑇𝑇 of phase, the magnetic field is reversed in direction𝐻𝐻𝐼𝐼 but does not undergo a phase change. The continuity of the magnetic field then requires that the transmitted field be twice as large.

Surface Impedance

As we have seen, the electric field is confined to a small depth at the conductor interface known as the skin depth. We define surface impedance as the ratio of the parallel component of electric field that gives rise to a current at the conductor surface,

= 𝐸𝐸∥ 𝑍𝑍𝑠𝑠 where is the surface current density. 𝐾𝐾𝑠𝑠

Assuming𝐾𝐾𝑠𝑠 that the current flows over the skin depth, one can write, for the current density, (assuming no reflection from the back of this depth)

= 0 −𝛾𝛾𝛾𝛾 Since the current density has been taken to decay𝐽𝐽 exponentially,𝐽𝐽 𝑒𝑒 we can extend the integration to infinity and get

0 = 0 = −𝛾𝛾𝛾𝛾 𝐽𝐽 𝐾𝐾𝑠𝑠 ∫ 𝐽𝐽 𝑒𝑒 𝑑𝑑𝑑𝑑 The current density at the surface can be written as . For a𝛾𝛾 good conductor, we have,

∥ = + 2 𝜎𝜎𝐸𝐸 (1 + ) 2

Thus 𝛾𝛾 �𝑖𝑖𝑖𝑖𝑖𝑖𝑖𝑖 𝜔𝜔 𝜇𝜇𝜇𝜇 ≈ 𝑖𝑖 �𝜔𝜔𝜔𝜔𝜔𝜔⁄

1 = = = (1 + ) = (1 + ) + 2 𝐸𝐸∥ 𝛾𝛾 𝜔𝜔𝜔𝜔 𝑍𝑍𝑠𝑠 � 𝑖𝑖 𝑖𝑖 ≡ 𝑅𝑅𝑠𝑠 𝑖𝑖𝑋𝑋𝑠𝑠 𝐾𝐾𝑠𝑠 𝜎𝜎 𝜎𝜎 𝜎𝜎𝜎𝜎

7 where the surface resistance and surface reactance are given by

𝑅𝑅𝑠𝑠 1 𝑋𝑋𝑠𝑠 = =

𝑅𝑅𝑠𝑠 𝑋𝑋𝑠𝑠 The current profile at the interface is as shown. 𝜎𝜎𝜎𝜎

Electromagnetic Waves

Lecture35: Electromagnetic Theory

Professor D. K. Ghosh, Physics Department, I.I.T., Bombay

Tutorial Assignment

1. A 2 GHz electromagnetic propagates in a non-magnetic medium having a relative permittivity of 20 and a conductivity of 3.85 S/m. Determine if the material is a good conductor or otherwise. Calculate the phase velocity of the wave, the propagation and attenuation constants, the skin depth and the intrinsic impedance.

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2. An electromagnetic wave with its electric field parallel to the plane of incidence is incident from vacuum onto the surface of a perfect conductor at an angle of incidence θ. Obtain an expression for the total electric and the magnetic field.

Solutions to Tutorial Assignments

1. One can see that = 2 × (2 × 109) × (20 × 8.85 × 10 12) = 2.22 S/m. The ratio of conductivity to is 1.73 which says it is neither a good −metal nor a good dielectric. The 𝜔𝜔𝜔𝜔 𝜋𝜋 propagation constant and the attenuation constant are given by, 𝜎𝜎 𝜔𝜔𝜔𝜔

𝛽𝛽 𝛼𝛼

2 = 1 + + 1 = 229.5 rad/m 2 2 2 𝜇𝜇𝜇𝜇 𝜎𝜎 𝛽𝛽 𝜔𝜔� ��� � 𝜔𝜔 𝜖𝜖 2 Np = 1 + 1 = 132.52 2 2 2 m 𝜇𝜇𝜇𝜇 𝜎𝜎 𝛼𝛼 𝜔𝜔� ��� − � 𝜔𝜔 𝜖𝜖 The intrinsic impedance is

= = 50.8 + 29.9 + 𝑖𝑖𝑖𝑖𝑖𝑖 𝜂𝜂 7� 𝑖𝑖 Ω The phase velocity is = 5.4 × 10 m𝜎𝜎/s. 𝑖𝑖𝑖𝑖𝑖𝑖 𝜔𝜔 1 The skin depth is =𝛽𝛽 = 7.5 mm

𝛿𝛿 𝛼𝛼 2. The case of p polarization is shown.

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Let the incident plane be y-z plane. Let us look at the magnetic field. We have, since both the incident and the reflected fields are in the same medium,

= = 𝐸𝐸𝑖𝑖 𝐸𝐸𝑟𝑟 Let us write the incident magnetic field as 𝜂𝜂 𝐻𝐻𝑖𝑖 𝐻𝐻𝑟𝑟 = ( sin cos ) The reflected magnetic field is given by −𝑖𝑖𝑖𝑖 𝑦𝑦 𝜃𝜃−𝑧𝑧 𝜃𝜃 𝐻𝐻��⃗𝑖𝑖 𝐻𝐻𝑖𝑖 𝑒𝑒 𝑥𝑥�

= ( sin + cos ) Since the tangential components of the electric−𝑖𝑖𝑖𝑖 𝑦𝑦field𝜃𝜃 is𝑧𝑧 continuous,𝜃𝜃 we have, 𝐻𝐻��⃗𝑟𝑟 𝐻𝐻𝑟𝑟 𝑒𝑒 𝑥𝑥� cos = cos As = , we have = , and consequently, = . Thus the total magnetic field can be 𝐸𝐸𝑖𝑖 𝜃𝜃 𝐸𝐸𝑟𝑟 𝜃𝜃𝑟𝑟 written as 𝜃𝜃𝑖𝑖 𝜃𝜃𝑟𝑟 𝐸𝐸𝑖𝑖 𝐸𝐸𝑟𝑟 𝐻𝐻𝑖𝑖 𝐻𝐻𝑟𝑟 = + = sin 2 cos( cos )

sin −𝑖𝑖𝑖𝑖𝑖𝑖 𝜃𝜃 𝐻𝐻=��⃗𝑡𝑡2 𝐻𝐻��⃗𝑖𝑖 𝐻𝐻��⃗𝑟𝑟 𝐻𝐻cos𝑖𝑖 𝑒𝑒( cos ) 𝛽𝛽𝛽𝛽 𝜃𝜃 𝑥𝑥� 𝐸𝐸𝑖𝑖 −𝑖𝑖𝑖𝑖𝑖𝑖 𝜃𝜃 The electric field has both y and𝑒𝑒 z components,𝛽𝛽 𝛽𝛽 𝜃𝜃 𝑥𝑥� 𝜂𝜂 = cos ( sin cos ) = sin −𝑖𝑖𝑖𝑖( 𝑦𝑦sin 𝜃𝜃−𝑧𝑧cos 𝜃𝜃) 𝐸𝐸𝑖𝑖𝑖𝑖 𝐸𝐸𝑖𝑖 𝜃𝜃𝑒𝑒 The reflected electric field also has both components,−𝑖𝑖𝑖𝑖 𝑦𝑦 𝜃𝜃−𝑧𝑧 𝜃𝜃 𝐸𝐸𝑖𝑖𝑖𝑖 𝐸𝐸𝑖𝑖 𝜃𝜃𝑒𝑒 = cos ( sin + cos ) = sin −(𝑖𝑖𝑖𝑖sin𝑦𝑦 +𝜃𝜃cos𝑧𝑧 ) 𝜃𝜃 𝐸𝐸𝑟𝑟𝑟𝑟 −𝐸𝐸𝑖𝑖 𝜃𝜃𝑒𝑒 Adding these two the total electric field, has the−𝑖𝑖 𝑖𝑖following𝑦𝑦 𝜃𝜃 𝑧𝑧 components,𝜃𝜃 𝐸𝐸𝑖𝑖𝑖𝑖 𝐸𝐸𝑖𝑖 𝜃𝜃𝑒𝑒

= cos sin cos cos = 2 cos −𝑖𝑖𝑖𝑖𝑖𝑖 sin𝜃𝜃 sin𝑖𝑖𝑖𝑖𝑖𝑖 ( 𝜃𝜃cos −)𝑖𝑖 𝑖𝑖𝑖𝑖 𝜃𝜃 𝐸𝐸𝑡𝑡𝑡𝑡 𝐸𝐸𝑖𝑖 𝜃𝜃𝑒𝑒 �𝑒𝑒 − 𝑒𝑒 � = sin −𝑖𝑖𝑖𝑖𝑖𝑖sin 𝜃𝜃 cos + cos 𝑖𝑖𝑖𝑖𝑖𝑖 𝜃𝜃𝑒𝑒 𝛽𝛽𝛽𝛽 𝜃𝜃 = 2 sin −𝑖𝑖sin𝑖𝑖𝑖𝑖 cos𝜃𝜃 ( 𝑖𝑖𝑖𝑖𝑖𝑖cos 𝜃𝜃) −𝑖𝑖𝑖𝑖𝑖𝑖 𝜃𝜃 𝐸𝐸𝑖𝑖𝑖𝑖 𝐸𝐸𝑖𝑖 𝜃𝜃𝑒𝑒 �𝑒𝑒 𝑒𝑒 � −𝑖𝑖𝑖𝑖𝑖𝑖 𝜃𝜃 𝐸𝐸𝑖𝑖 𝜃𝜃𝑒𝑒 𝛽𝛽𝛽𝛽 𝜃𝜃

3.

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Electromagnetic Waves

Lecture35: Electromagnetic Theory

Professor D. K. Ghosh, Physics Department, I.I.T., Bombay

Self Assessment Questions

1. For electromagnetic wave propagation inside a good conductor, show that the electric and the magnetic fields are out of phase by 450. 2. A 2 kHz electromagnetic propagates in a non-magnetic medium having a relative permittivity of 20 and a conductivity of 3.85 S/m. Determine if the material is a good conductor or otherwise. Calculate the phase velocity of the wave, the propagation and attenuation constants, the skin depth and the intrinsic impedance. 3. An electromagnetic wave with its electric field perpendicular to the plane of incidence is incident from vacuum onto the surface of a perfect conductor at an angle of incidence θ. Obtain an expression for the total electric and the magnetic field.

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Solutions to Self Assessment Questions

1. From the text, we see that the ratio of electric field to magnetic field is given by

= + 𝐸𝐸𝑥𝑥 𝑖𝑖𝑖𝑖𝑖𝑖 𝑦𝑦 � 𝐻𝐻 𝜎𝜎 𝑖𝑖𝑖𝑖𝑖𝑖= 4 For a good conductor, we can approximate this by 𝑖𝑖𝑖𝑖 . 𝑖𝑖𝑖𝑖𝑖𝑖 𝜔𝜔𝜔𝜔 3 12 6 2. One can see that = 2 × (2 × 10 ) × (20 × 8.�85𝜎𝜎× 10𝑒𝑒 �) =𝜎𝜎 2.22 × 10 S/m. The ratio of conductivity to is 1.73 × 106 which says it is r a good− metal. The propagation− constant 𝜔𝜔𝜔𝜔 𝜋𝜋 and the attenuation constant are given by, 𝜎𝜎 𝜔𝜔𝜔𝜔

𝛽𝛽 𝛼𝛼

= = 0.176 rad/m 2 𝜔𝜔𝜔𝜔𝜔𝜔 � 𝛽𝛽 = = 0.176 Np/m 2 𝜔𝜔𝜔𝜔𝜔𝜔 𝛼𝛼 � The intrinsic impedance is

= = 0.045(1 + ) + 𝑖𝑖𝑖𝑖𝑖𝑖 𝜂𝜂 �4 𝑖𝑖 Ω The phase velocity is = 7.14 × 10 𝜎𝜎m/s.𝑖𝑖𝑖𝑖 𝑖𝑖 𝜔𝜔 1 The skin depth is =𝛽𝛽 = 5.68 m. 3. The direction of electric and magnetic field for s polarization is as shown below. 𝛿𝛿 𝛼𝛼 Let the incident plane be y-z plane. The incident electric field is

taken along the x direction and is given by

= ( sin cos ) = −𝑖𝑖𝑖𝑖 𝑦𝑦 ( sin𝜃𝜃−𝑧𝑧+ cos𝜃𝜃 ) 𝐸𝐸�⃗𝑖𝑖 𝐸𝐸𝑖𝑖 𝑒𝑒 𝑥𝑥� −𝑖𝑖𝑖𝑖 𝑦𝑦 𝜃𝜃 𝑧𝑧 𝜃𝜃 The minus sign comes𝐸𝐸�⃗𝑟𝑟 because−𝐸𝐸𝑖𝑖 𝑒𝑒 at z=0, for any 𝑥𝑥�y, the tangential component of the electric field must be zero.

The total electric field is along the x direction and is given by the sum of the above,

= 2 sin sin( cos ) −𝑖𝑖𝑖𝑖𝑖𝑖 𝜃𝜃 𝐸𝐸�⃗𝑡𝑡 𝑖𝑖𝑖𝑖𝑖𝑖 𝑒𝑒 𝛽𝛽𝛽𝛽 𝜃𝜃 𝑥𝑥� 12 which is a travelling wave in the y direction but a standing wave in the z direction. Since the wave propagates in vacuum, we have,

= = 𝐸𝐸𝑖𝑖 𝐸𝐸𝑟𝑟 𝜂𝜂 The magnetic field has both y and z components. 𝐻𝐻 𝑖𝑖 𝐻𝐻𝑟𝑟

= + = cos ( sin cos ) cos ( sin + cos ) = 2 cos sin cos ( cos−𝑖𝑖𝑖𝑖 𝑦𝑦) 𝜃𝜃−𝑧𝑧 𝜃𝜃 −𝑖𝑖𝑖𝑖 𝑦𝑦 𝜃𝜃 𝑧𝑧 𝜃𝜃 𝐻𝐻𝑡𝑡𝑡𝑡 𝐻𝐻𝑖𝑖𝑖𝑖 𝐻𝐻𝑟𝑟𝑟𝑟 −𝐻𝐻𝑖𝑖 𝜃𝜃 𝑒𝑒 − 𝐻𝐻𝑖𝑖 𝜃𝜃𝑒𝑒 −𝑖𝑖𝑖𝑖𝑖𝑖 sin 𝜃𝜃 = −2𝐻𝐻𝑖𝑖 cos𝜃𝜃𝑒𝑒 cos (𝛽𝛽𝛽𝛽 cos𝜃𝜃) 𝐸𝐸𝑖𝑖 −𝑖𝑖𝑖𝑖𝑖𝑖 𝜃𝜃 −= +𝜃𝜃𝑒𝑒 = sin 𝛽𝛽 𝛽𝛽 ( 𝜃𝜃sin cos ) + sin ( sin + cos ) 𝜂𝜂 = 2 cos sin sin ( cos−𝑖𝑖𝑖𝑖 𝑦𝑦) 𝜃𝜃−𝑧𝑧 𝜃𝜃 −𝑖𝑖𝑖𝑖 𝑦𝑦 𝜃𝜃 𝑧𝑧 𝜃𝜃 𝐻𝐻𝑡𝑡𝑡𝑡 𝐻𝐻𝑖𝑖𝑖𝑖 𝐻𝐻𝑟𝑟𝑟𝑟 −𝐻𝐻𝑖𝑖 𝜃𝜃 𝑒𝑒 𝐻𝐻𝑖𝑖 𝜃𝜃 𝑒𝑒 −𝑖𝑖𝑖𝑖𝑖𝑖sin 𝜃𝜃 = 2𝑖𝑖𝐻𝐻𝑖𝑖cos 𝜃𝜃𝑒𝑒 sin ( 𝛽𝛽𝛽𝛽cos 𝜃𝜃) 𝐸𝐸𝑖𝑖 −𝑖𝑖𝑖𝑖𝑖𝑖 𝜃𝜃 𝜃𝜃𝑒𝑒 𝛽𝛽𝛽𝛽 𝜃𝜃 𝜂𝜂

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