* Your assessment is very important for improving the workof artificial intelligence, which forms the content of this project
Download Electromagnetic Waves
Survey
Document related concepts
History of quantum field theory wikipedia , lookup
Magnetic monopole wikipedia , lookup
History of electromagnetic theory wikipedia , lookup
Introduction to gauge theory wikipedia , lookup
Electrostatics wikipedia , lookup
Electromagnet wikipedia , lookup
Maxwell's equations wikipedia , lookup
Lorentz force wikipedia , lookup
Time in physics wikipedia , lookup
Field (physics) wikipedia , lookup
Superconductivity wikipedia , lookup
Theoretical and experimental justification for the Schrödinger equation wikipedia , lookup
Transcript
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. �⃗ and 𝑘𝑘�⃗ As the medium is linear and the propagation takes place in the infinite medium, the vectors 𝐸𝐸�⃗ , 𝐻𝐻 are still mutually perpendicular. We take the electric field along the x direction, the magnetic field along 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, 𝐽𝐽⃗ = 𝜎𝜎𝐸𝐸�⃗ . Consider the pair of curl equations of Maxwell. �⃗ �⃗ 𝜕𝜕𝐵𝐵 𝜕𝜕𝐻𝐻 = −μ 𝜕𝜕𝜕𝜕 𝜕𝜕𝜕𝜕 𝜕𝜕𝐸𝐸�⃗ 𝜕𝜕𝐸𝐸�⃗ �⃗ = 𝐽𝐽⃗ + 𝜖𝜖 𝛻𝛻 × 𝐻𝐻 = 𝜎𝜎𝐸𝐸�⃗ + 𝜖𝜖 𝜕𝜕𝜕𝜕 𝜕𝜕𝜕𝜕 𝛻𝛻 × 𝐸𝐸�⃗ = − �⃗ and 𝑘𝑘�⃗ to be respectively in x, y and z direction. We the nhave, Let us take 𝐸𝐸�⃗ , 𝐻𝐻 i.e., and i.e. �𝛻𝛻 × 𝐸𝐸�⃗ �𝑦𝑦 = 𝜕𝜕𝐻𝐻𝑦𝑦 𝜕𝜕𝐸𝐸𝑥𝑥 = −μ 𝜕𝜕𝜕𝜕 𝜕𝜕𝜕𝜕 𝜕𝜕𝐻𝐻𝑦𝑦 𝜕𝜕𝐸𝐸𝑥𝑥 +μ =0 𝜕𝜕𝜕𝜕 𝜕𝜕𝜕𝜕 �⃗� = − �𝛻𝛻 × 𝐻𝐻 𝑥𝑥 (1) 𝜕𝜕𝐻𝐻𝑦𝑦 𝜕𝜕𝐸𝐸𝑥𝑥 = 𝜎𝜎𝐸𝐸𝑥𝑥 + 𝜖𝜖 𝜕𝜕𝜕𝜕 𝜕𝜕𝜕𝜕 𝜕𝜕𝐻𝐻𝑦𝑦 𝜕𝜕𝐸𝐸𝑥𝑥 + 𝜎𝜎𝐸𝐸𝑥𝑥 + 𝜖𝜖 = 0 (2) 𝜕𝜕𝜕𝜕 𝜕𝜕𝜕𝜕 1 We take the time variation to be harmonic ( ~𝑒𝑒 𝑖𝑖𝑖𝑖𝑖𝑖 ) so that the time derivative is equivalent to a multiplication by 𝑖𝑖𝑖𝑖. The pair of equations (1) and (2) can then be written as 𝜕𝜕𝐸𝐸𝑥𝑥 + 𝑖𝑖μωHy = 0 𝜕𝜕𝜕𝜕 𝜕𝜕𝐻𝐻𝑦𝑦 + 𝜎𝜎𝐸𝐸𝑥𝑥 + 𝑖𝑖𝑖𝑖𝑖𝑖𝐸𝐸𝑥𝑥 = 0 𝜕𝜕𝜕𝜕 We can solve this pair of coupled equations by taking a derivative of either of the equations with respect to z and substituting the other into it, ∂Hy 𝜕𝜕 2 𝐸𝐸𝑥𝑥 𝜕𝜕 2 𝐸𝐸𝑥𝑥 + 𝑖𝑖μω = − 𝑖𝑖𝑖𝑖𝑖𝑖(𝜎𝜎 + 𝑖𝑖𝑖𝑖𝑖𝑖)𝐸𝐸𝑥𝑥 = 0 𝜕𝜕𝑧𝑧 2 𝜕𝜕𝜕𝜕 𝜕𝜕𝑧𝑧 2 Define, a complex constant γ through in terms of which we have, In an identical fashion, we get 𝛾𝛾 2 = 𝑖𝑖𝑖𝑖𝑖𝑖(𝜎𝜎 + 𝑖𝑖𝑖𝑖𝑖𝑖) 𝜕𝜕 2 𝐸𝐸𝑥𝑥 − 𝛾𝛾 2 𝐸𝐸𝑥𝑥 = 0 𝜕𝜕𝑧𝑧 2 (3) 𝜕𝜕 2 𝐻𝐻𝑦𝑦 − 𝛾𝛾 2 𝐻𝐻𝑦𝑦 = 0 𝜕𝜕𝑧𝑧 2 (4) Solutions of (3) and (4) are well known and are expressed in terms of hyperbolic functions, 𝐸𝐸𝑥𝑥 = 𝐴𝐴 cosh(𝛾𝛾𝛾𝛾) + 𝐵𝐵 sinh(𝛾𝛾𝛾𝛾) 𝐻𝐻𝑦𝑦 = 𝐶𝐶 cosh(𝛾𝛾𝛾𝛾) + 𝐷𝐷 sinh(𝛾𝛾𝛾𝛾) where A, B, C and D are constants to be determined. If the values of the electric field at z=0 is 𝐸𝐸0 and that of the magnetic field at z=0 is 𝐻𝐻0 , we have 𝐴𝐴 = 𝐸𝐸0 and 𝐶𝐶 = 𝐻𝐻0 . In order to determine the constants B and D, let us return back to the original first order equations (1) and (2) 𝜕𝜕𝐸𝐸𝑥𝑥 + 𝑖𝑖μωHy = 0 𝜕𝜕𝜕𝜕 𝜕𝜕𝐻𝐻𝑦𝑦 + 𝜎𝜎𝐸𝐸𝑥𝑥 + 𝑖𝑖𝑖𝑖𝑖𝑖𝐸𝐸𝑥𝑥 = 0 𝜕𝜕𝜕𝜕 Substituting the solutions for E and H 𝛾𝛾𝐸𝐸0 sinh(𝛾𝛾𝛾𝛾) + 𝐵𝐵𝐵𝐵 cosh(𝛾𝛾𝛾𝛾) + 𝑖𝑖𝑖𝑖𝑖𝑖(H0 cosh(𝛾𝛾𝛾𝛾) + 𝐷𝐷 sinh(𝛾𝛾𝛾𝛾)) = 0 This equation must remain valid for all values of z, which is possible if the coefficients of 𝑠𝑠𝑠𝑠𝑠𝑠ℎ and cosh terms are separately equated to zero, 𝐸𝐸0 𝛾𝛾 + 𝑖𝑖𝑖𝑖𝑖𝑖𝑖𝑖 = 0 𝐵𝐵𝐵𝐵 + 𝑖𝑖𝑖𝑖𝑖𝑖𝐻𝐻0 = 0 The former gives, 2 𝐷𝐷 = − 𝛾𝛾 𝐸𝐸 𝑖𝑖𝑖𝑖𝑖𝑖 0 =− �𝑖𝑖𝑖𝑖𝑖𝑖(𝜎𝜎 + 𝑖𝑖𝑖𝑖𝑖𝑖) 𝑖𝑖𝑖𝑖𝑖𝑖 =− 𝐸𝐸0 𝜂𝜂 𝜎𝜎 + 𝑖𝑖𝑖𝑖𝑖𝑖 𝐸𝐸0 𝑖𝑖𝑖𝑖𝑖𝑖 = −� where 𝑖𝑖𝑖𝑖𝑖𝑖 𝜎𝜎 + 𝑖𝑖𝑖𝑖𝑖𝑖 𝜂𝜂 = � Likewise, we get, 𝐵𝐵 = −𝜂𝜂𝐻𝐻0 Substituting these, our solutions for the E and H become, 𝐸𝐸𝑥𝑥 = 𝐸𝐸0 cosh(𝛾𝛾𝛾𝛾) − 𝜂𝜂𝐻𝐻0 sinh(𝛾𝛾𝛾𝛾) 𝐸𝐸0 𝐻𝐻𝑦𝑦 = 𝐻𝐻0 cosh(𝛾𝛾𝛾𝛾) − sinh(𝛾𝛾𝛾𝛾) 𝜂𝜂 The wave is propagating in the z direction. Let us evaluate the fields when the wave has reached 𝑧𝑧 = 𝑙𝑙, 𝐸𝐸𝑥𝑥 = 𝐸𝐸0 cosh(𝛾𝛾𝛾𝛾) − 𝜂𝜂𝐻𝐻0 sinh(𝛾𝛾𝛾𝛾) 𝐸𝐸0 𝐻𝐻𝑦𝑦 = 𝐻𝐻0 cosh(𝛾𝛾𝛾𝛾) − sinh(𝛾𝛾𝛾𝛾) 𝜂𝜂 If 𝑙𝑙 is large, we can approximate 𝑒𝑒 𝛾𝛾𝛾𝛾 sinh(𝛾𝛾𝛾𝛾) ≈ cosh(𝛾𝛾𝛾𝛾) = 2 we then have, 𝑒𝑒 𝛾𝛾𝛾𝛾 𝐸𝐸𝑥𝑥 = (𝐸𝐸0 − 𝜂𝜂𝐻𝐻0 ) 2 𝐸𝐸0 𝑒𝑒 𝛾𝛾𝛾𝛾 𝐻𝐻𝑦𝑦 = (𝐻𝐻0 − ) 𝜂𝜂 2 The ratio of the magnitudes of the electric field to magnetic field is defined as the “characteristic impedance” of the wave 𝑖𝑖𝑖𝑖𝑖𝑖 𝐸𝐸𝑥𝑥 � � = 𝜂𝜂 = � 𝜎𝜎 + 𝑖𝑖𝑖𝑖𝑖𝑖 𝐻𝐻𝑦𝑦 Suppose we have lossless medium, σ=0, i.e. for a perfect conductor, the characteristic impedance is 3 𝜇𝜇 𝜂𝜂 = � 𝜖𝜖 If the medium is vacuum, 𝜇𝜇 = 𝜇𝜇0 = 4𝜋𝜋 × 10−7 N⁄A2 and 𝜖𝜖 = 𝜖𝜖0 = 8.85 × 10−12 C 2 ⁄N − m2 gives 𝜂𝜂 ≈ 377 𝛺𝛺. The characteristic impedance, as the name suggests, has the dimension of resistance. In this case, 𝛾𝛾 = 𝑖𝑖 √𝜇𝜇𝜇𝜇. Let us look at the full three dimensional version of the propagation in a conductor. Once again, we start with the two curl equations, 𝛻𝛻 × 𝐸𝐸�⃗ = −μ �⃗ 𝜕𝜕𝐻𝐻 𝜕𝜕𝜕𝜕 �⃗ = 𝜎𝜎𝐸𝐸�⃗ + 𝜖𝜖 𝛻𝛻 × 𝐻𝐻 Take a curl of both sides of the first equation, 𝜕𝜕𝐸𝐸�⃗ 𝜕𝜕𝜕𝜕 𝛻𝛻 × �𝛻𝛻 × 𝐸𝐸�⃗ � = 𝛻𝛻�𝛻𝛻 ⋅ 𝐸𝐸�⃗ � − 𝛻𝛻 2 𝐸𝐸�⃗ = −μ �⃗) 𝜕𝜕(𝛻𝛻 × 𝐻𝐻 𝜕𝜕𝜕𝜕 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�⃗ = μ We take the propagating solutions to be so that the above equation becomes, = 𝜎𝜎𝜎𝜎 𝜕𝜕 𝜕𝜕𝐸𝐸�⃗ �𝜎𝜎𝐸𝐸�⃗ + 𝜖𝜖 � 𝜕𝜕𝜕𝜕 𝜕𝜕𝜕𝜕 𝜕𝜕𝐸𝐸�⃗ 𝜕𝜕 2 𝐸𝐸�⃗ + 𝜎𝜎𝜎𝜎 2 𝜕𝜕𝜕𝜕 𝜕𝜕𝑡𝑡 �⃗ 𝐸𝐸�⃗ = 𝐸𝐸�⃗0 𝑒𝑒 𝑖𝑖�𝑘𝑘 ⋅𝑟𝑟⃗−𝜔𝜔𝜔𝜔 � 𝑘𝑘 2 𝐸𝐸�⃗ = (𝑖𝑖𝑖𝑖𝑖𝑖𝑖𝑖 + 𝜔𝜔2 𝜇𝜇𝜇𝜇)𝐸𝐸�⃗ so that we have, the complex propagation constant to be given by so that 𝑘𝑘 2 = 𝑖𝑖𝑖𝑖𝑖𝑖𝑖𝑖 + 𝜔𝜔2 𝜇𝜇𝜇𝜇 𝑘𝑘 = �𝜔𝜔𝜔𝜔(𝜔𝜔𝜔𝜔 + 𝑖𝑖𝑖𝑖) k is complex and its real and imaginary parts can be separated by standard algebra, 4 we have 1/2 𝜎𝜎 2 𝜇𝜇𝜇𝜇 𝑘𝑘 = 𝜔𝜔� ��1 + �1 + 2 2 � 𝜔𝜔 𝜖𝜖 2 1/2 𝜎𝜎 2 + 𝑖𝑖 ��1 + 2 2 − 1� 𝜔𝜔 𝜖𝜖 � Thus the propagation vector β and the attenuation factor α are given by 𝜎𝜎 2 𝜇𝜇𝜇𝜇 𝛽𝛽 = 𝜔𝜔� ���1 + 2 2 + 1� 𝜔𝜔 𝜖𝜖 2 𝛼𝛼 = 𝜔𝜔� The ration 𝜎𝜎 𝜔𝜔𝜔𝜔 𝜎𝜎 2 𝜇𝜇𝜇𝜇 ���1 + 2 2 − 1� 𝜔𝜔 𝜖𝜖 2 determines whether a material is a good conductor or otherwise. Consider a good conductor for which 𝜎𝜎 ≫ 𝜔𝜔𝜔𝜔. For this case, we have, 𝛽𝛽 = 𝛼𝛼 = � The speed of electromagnetic wave is given by 𝜔𝜔𝜔𝜔𝜔𝜔 2 𝑣𝑣 = 𝜔𝜔 2𝜔𝜔 = � 𝛽𝛽 𝜎𝜎𝜎𝜎 𝛿𝛿 = 1 2 = � 𝛼𝛼 𝜔𝜔𝜔𝜔𝜔𝜔 The electric field amplitude diminishes with distance as 𝑒𝑒 −𝛼𝛼𝛼𝛼 . The distance to which the field penetrates before its amplitude diminishes be a factor 𝑒𝑒 −1 is known as the “skin depth” , which is given by 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 𝛺𝛺 −1 m−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 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. 5 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 𝐻𝐻𝐼𝐼 + 𝐻𝐻𝑅𝑅 = 𝐻𝐻𝑇𝑇 The relationships between the magnetic field and the electric field are given by 𝐻𝐻𝐼𝐼 = 𝐸𝐸𝐼𝐼 𝜂𝜂1 𝐻𝐻𝑇𝑇 = 𝐸𝐸𝑇𝑇 𝜂𝜂2 𝐻𝐻𝑅𝑅 = − 𝐸𝐸𝑅𝑅 𝜂𝜂1 the minus sign in the second relation comes because of the propagation direction having been reversed on reflection. Solving these, we get, 𝐸𝐸𝑅𝑅 𝜂𝜂2 − 𝜂𝜂1 = 𝐸𝐸𝐼𝐼 𝜂𝜂2 + 𝜂𝜂1 2𝜂𝜂2 𝐸𝐸𝑇𝑇 𝐸𝐸𝑅𝑅 = = 𝐸𝐸𝐼𝐼 𝐸𝐸𝐼𝐼 𝜂𝜂2 + 𝜂𝜂1 The magnetic field expressions are given by interchanging 𝜂𝜂1 and 𝜂𝜂2 in the above expressions. E 6 Let us look at consequence of this. Consider a good conductor such as copper. We can see that 𝜂𝜂2 is a small complex number. For instance, taking the wave frequency to be 1 MHz and substituting conductivity of Cu to be 6 × 106 𝛺𝛺−1 𝑚𝑚−1 , we can calculate 𝜂𝜂2 to be approximately (1 + 𝑖𝑖) × 2.57 × 10−4 𝛺𝛺 whereas the vacuum impedance 𝜂𝜂1 = 377 𝛺𝛺. This implies 𝐸𝐸𝑅𝑅 𝜂𝜂2 − 𝜂𝜂1 = ≈ −1 𝐸𝐸𝐼𝐼 𝜂𝜂2 + 𝜂𝜂1 which shows that a good metal is also a good reflector. On the other hand, if we calculate the transmission coefficient we find it to be substantially reduced, being only about (1 + 𝑖𝑖) × 1−6 . For the transmitted magnetic field, the ratio 𝐻𝐻𝑇𝑇 𝐻𝐻𝐼𝐼 is approximately +2. Though E is reflected with a change of phase, the magnetic field is reversed in direction 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, where 𝐾𝐾𝑠𝑠 is the surface current density. 𝑍𝑍𝑠𝑠 = 𝐸𝐸∥ 𝐾𝐾𝑠𝑠 Assuming that the current flows over the skin depth, one can write, for the current density, (assuming no reflection from the back of this depth) 𝐽𝐽 = 𝐽𝐽0 𝑒𝑒 −𝛾𝛾𝛾𝛾 Since the current density has been taken to decay exponentially, we can extend the integration to infinity and get 𝐾𝐾𝑠𝑠 = ∫ 𝐽𝐽0 𝑒𝑒 −𝛾𝛾𝛾𝛾 𝑑𝑑𝑑𝑑 = 𝐽𝐽0 𝛾𝛾 The current density at the surface can be written as 𝜎𝜎𝐸𝐸∥ . For a good conductor, we have, 𝛾𝛾 = �𝑖𝑖𝑖𝑖𝑖𝑖𝑖𝑖 + 𝜔𝜔 2 𝜇𝜇𝜇𝜇 ≈ (1 + 𝑖𝑖)�𝜔𝜔𝜔𝜔𝜔𝜔⁄2 Thus 𝑍𝑍𝑠𝑠 = 𝐸𝐸∥ 𝛾𝛾 1 𝜔𝜔𝜔𝜔 (1 + 𝑖𝑖) ≡ 𝑅𝑅𝑠𝑠 + 𝑖𝑖𝑋𝑋𝑠𝑠 = = � (1 + 𝑖𝑖) = 𝐾𝐾𝑠𝑠 𝜎𝜎 𝜎𝜎𝜎𝜎 2𝜎𝜎 7 where the surface resistance 𝑅𝑅𝑠𝑠 and surface reactance 𝑋𝑋𝑠𝑠 are given by 𝑅𝑅𝑠𝑠 = The current profile at the interface is as shown. 1 = 𝑋𝑋𝑠𝑠 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. 8 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 𝜔𝜔𝜔𝜔 = 2𝜋𝜋 × (2 × 109 ) × (20 × 8.85 × 10−12 ) = 2.22 S/m. The ratio of conductivity 𝜎𝜎 to 𝜔𝜔𝜔𝜔 is 1.73 which says it is neither a good metal nor a good dielectric. The propagation constant 𝛽𝛽 and the attenuation constant 𝛼𝛼 are given by, 𝜎𝜎 2 𝜇𝜇𝜇𝜇 𝛽𝛽 = 𝜔𝜔� ���1 + 2 2 + 1� = 229.5 rad/m 𝜔𝜔 𝜖𝜖 2 The intrinsic impedance is 𝜎𝜎 2 Np 𝜇𝜇𝜇𝜇 𝛼𝛼 = 𝜔𝜔� ���1 + 2 2 − 1� = 132.52 m 𝜔𝜔 𝜖𝜖 2 The phase velocity is 𝜔𝜔 𝛽𝛽 1 The skin depth is 𝛿𝛿 = 𝛼𝛼 𝜂𝜂 = � 𝑖𝑖𝑖𝑖𝑖𝑖 = 50.8 + 29.9 𝑖𝑖 Ω 𝜎𝜎 + 𝑖𝑖𝑖𝑖𝑖𝑖 = 5.4 × 107 m/s. = 7.5 mm 2. The case of p polarization is shown. 9 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, 𝐸𝐸𝑟𝑟 𝐸𝐸𝑖𝑖 = = 𝜂𝜂 𝐻𝐻𝑖𝑖 𝐻𝐻𝑟𝑟 Let us write the incident magnetic field as �⃗𝑖𝑖 = 𝐻𝐻𝑖𝑖 𝑒𝑒 −𝑖𝑖𝑖𝑖 (𝑦𝑦sin 𝜃𝜃−𝑧𝑧 cos 𝜃𝜃) 𝑥𝑥� 𝐻𝐻 The reflected magnetic field is given by �⃗𝑟𝑟 = 𝐻𝐻𝑟𝑟 𝑒𝑒 −𝑖𝑖𝑖𝑖 (𝑦𝑦sin 𝜃𝜃+𝑧𝑧 cos 𝜃𝜃) 𝑥𝑥� 𝐻𝐻 Since the tangential components of the electric field is continuous, we have, 𝐸𝐸𝑖𝑖 cos 𝜃𝜃 = 𝐸𝐸𝑟𝑟 cos 𝜃𝜃𝑟𝑟 As 𝜃𝜃𝑖𝑖 = 𝜃𝜃𝑟𝑟 , we have 𝐸𝐸𝑖𝑖 = 𝐸𝐸𝑟𝑟 , and consequently, 𝐻𝐻𝑖𝑖 = 𝐻𝐻𝑟𝑟 . Thus the total magnetic field can be written as �⃗𝑖𝑖 + 𝐻𝐻 �⃗𝑡𝑡 = 𝐻𝐻 �⃗𝑟𝑟 = 𝐻𝐻𝑖𝑖 𝑒𝑒 −𝑖𝑖𝑖𝑖𝑖𝑖 sin 𝜃𝜃 2 cos(𝛽𝛽𝛽𝛽 cos 𝜃𝜃)𝑥𝑥� 𝐻𝐻 𝐸𝐸𝑖𝑖 = 2 𝑒𝑒 −𝑖𝑖𝑖𝑖𝑖𝑖 sin 𝜃𝜃 cos(𝛽𝛽𝛽𝛽 cos 𝜃𝜃)𝑥𝑥� 𝜂𝜂 The electric field has both y and z components, 𝐸𝐸𝑖𝑖𝑖𝑖 = 𝐸𝐸𝑖𝑖 cos 𝜃𝜃𝑒𝑒 −𝑖𝑖𝑖𝑖 (𝑦𝑦sin 𝜃𝜃−𝑧𝑧 cos 𝜃𝜃) 𝐸𝐸𝑖𝑖𝑖𝑖 = 𝐸𝐸𝑖𝑖 sin 𝜃𝜃𝑒𝑒 −𝑖𝑖𝑖𝑖 (𝑦𝑦sin 𝜃𝜃−𝑧𝑧 cos 𝜃𝜃) The reflected electric field also has both components, 𝐸𝐸𝑟𝑟𝑟𝑟 = −𝐸𝐸𝑖𝑖 cos 𝜃𝜃𝑒𝑒 −𝑖𝑖𝑖𝑖 (𝑦𝑦sin 𝜃𝜃+𝑧𝑧 cos 𝜃𝜃 ) 𝐸𝐸𝑖𝑖𝑖𝑖 = 𝐸𝐸𝑖𝑖 sin 𝜃𝜃𝑒𝑒 −𝑖𝑖𝑖𝑖 (𝑦𝑦sin 𝜃𝜃 +𝑧𝑧 cos 𝜃𝜃 ) Adding these two the total electric field, has the following components, 𝐸𝐸𝑡𝑡𝑡𝑡 = 𝐸𝐸𝑖𝑖 cos 𝜃𝜃𝑒𝑒 −𝑖𝑖𝑖𝑖𝑖𝑖 sin 𝜃𝜃 �𝑒𝑒 𝑖𝑖𝑖𝑖𝑖𝑖 cos 𝜃𝜃 − 𝑒𝑒 −𝑖𝑖𝑖𝑖𝑖𝑖 cos 𝜃𝜃 � = 2𝑖𝑖𝑖𝑖𝑖𝑖 cos 𝜃𝜃𝑒𝑒 −𝑖𝑖𝑖𝑖𝑖𝑖 sin 𝜃𝜃 sin(𝛽𝛽𝛽𝛽 cos 𝜃𝜃) 𝐸𝐸𝑖𝑖𝑖𝑖 = 𝐸𝐸𝑖𝑖 sin 𝜃𝜃𝑒𝑒 −𝑖𝑖𝑖𝑖𝑖𝑖 sin 𝜃𝜃 �𝑒𝑒 𝑖𝑖𝑖𝑖𝑖𝑖 cos 𝜃𝜃 + 𝑒𝑒 −𝑖𝑖𝑖𝑖𝑖𝑖 cos 𝜃𝜃 � = 2𝐸𝐸𝑖𝑖 sin 𝜃𝜃𝑒𝑒 −𝑖𝑖𝑖𝑖𝑖𝑖 sin 𝜃𝜃 cos(𝛽𝛽𝛽𝛽 cos 𝜃𝜃) 3. 10 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. 11 Solutions to Self Assessment Questions 1. From the text, we see that the ratio of electric field to magnetic field is given by 𝑖𝑖𝑖𝑖𝑖𝑖 𝐸𝐸𝑥𝑥 = � 𝜎𝜎 + 𝑖𝑖𝑖𝑖𝑖𝑖 𝐻𝐻𝑦𝑦 For a good conductor, we can approximate this by � 𝑖𝑖𝑖𝑖𝑖𝑖 𝜎𝜎 𝑖𝑖𝑖𝑖 𝜔𝜔𝜔𝜔 = 𝑒𝑒 4 � 𝜎𝜎 . 2. One can see that 𝜔𝜔𝜔𝜔 = 2𝜋𝜋 × (2 × 103 ) × (20 × 8.85 × 10−12 ) = 2.22 × 10−6 S/m. The ratio of conductivity 𝜎𝜎 to 𝜔𝜔𝜔𝜔 is 1.73 × 106 which says it is r a good metal. The propagation constant 𝛽𝛽 and the attenuation constant 𝛼𝛼 are given by, 𝜔𝜔𝜔𝜔𝜔𝜔 𝛽𝛽 = � = 0.176 rad/m 2 𝜔𝜔𝜔𝜔𝜔𝜔 = 0.176 Np/m 𝛼𝛼 = � 2 The intrinsic impedance is The phase velocity is 𝜔𝜔 𝛽𝛽 1 The skin depth is 𝛿𝛿 = 𝛼𝛼 𝑖𝑖𝑖𝑖𝑖𝑖 = 0.045(1 + 𝑖𝑖 ) Ω 𝜎𝜎 + 𝑖𝑖𝑖𝑖𝑖𝑖 𝜂𝜂 = � = 7.14 × 104 m/s. = 5.68 m. 3. The direction of electric and magnetic field for s polarization is as shown below. Let the incident plane be y-z plane. The incident electric field is taken along the x direction and is given by 𝐸𝐸�⃗𝑖𝑖 = 𝐸𝐸𝑖𝑖 𝑒𝑒 −𝑖𝑖𝑖𝑖 (𝑦𝑦 sin 𝜃𝜃−𝑧𝑧 cos 𝜃𝜃) 𝑥𝑥� 𝐸𝐸�⃗𝑟𝑟 = −𝐸𝐸𝑖𝑖 𝑒𝑒 −𝑖𝑖𝑖𝑖 (𝑦𝑦 sin 𝜃𝜃 +𝑧𝑧 cos 𝜃𝜃 ) 𝑥𝑥� The minus sign comes because at z=0, for any 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, 𝐸𝐸�⃗𝑡𝑡 = 2𝑖𝑖𝑖𝑖𝑖𝑖 𝑒𝑒 −𝑖𝑖𝑖𝑖𝑖𝑖 sin 𝜃𝜃 sin(𝛽𝛽𝛽𝛽 cos 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, The magnetic field has both y and z components. 𝐸𝐸𝑖𝑖 𝐸𝐸𝑟𝑟 = = 𝜂𝜂 𝐻𝐻𝑖𝑖 𝐻𝐻𝑟𝑟 𝐻𝐻𝑡𝑡𝑡𝑡 = 𝐻𝐻𝑖𝑖𝑖𝑖 + 𝐻𝐻𝑟𝑟𝑟𝑟 = −𝐻𝐻𝑖𝑖 cos 𝜃𝜃 𝑒𝑒 −𝑖𝑖𝑖𝑖 (𝑦𝑦 sin 𝜃𝜃 −𝑧𝑧 cos 𝜃𝜃) − 𝐻𝐻𝑖𝑖 cos 𝜃𝜃𝑒𝑒 −𝑖𝑖𝑖𝑖 (𝑦𝑦 sin 𝜃𝜃 +𝑧𝑧 cos 𝜃𝜃 ) = −2𝐻𝐻𝑖𝑖 cos 𝜃𝜃𝑒𝑒 −𝑖𝑖𝑖𝑖𝑖𝑖 sin 𝜃𝜃 cos (𝛽𝛽𝛽𝛽 cos 𝜃𝜃) 𝐸𝐸𝑖𝑖 = −2 cos 𝜃𝜃𝑒𝑒 −𝑖𝑖𝑖𝑖𝑖𝑖 sin 𝜃𝜃 cos (𝛽𝛽𝛽𝛽 cos 𝜃𝜃) 𝜂𝜂 𝐻𝐻𝑡𝑡𝑡𝑡 = 𝐻𝐻𝑖𝑖𝑖𝑖 + 𝐻𝐻𝑟𝑟𝑟𝑟 = −𝐻𝐻𝑖𝑖 sin 𝜃𝜃 𝑒𝑒 −𝑖𝑖𝑖𝑖 (𝑦𝑦 sin 𝜃𝜃−𝑧𝑧 cos 𝜃𝜃) + 𝐻𝐻𝑖𝑖 sin 𝜃𝜃 𝑒𝑒 −𝑖𝑖𝑖𝑖 (𝑦𝑦 sin 𝜃𝜃 +𝑧𝑧 cos 𝜃𝜃 ) = 2𝑖𝑖𝐻𝐻𝑖𝑖 cos 𝜃𝜃𝑒𝑒 −𝑖𝑖𝑖𝑖𝑖𝑖 sin 𝜃𝜃 sin (𝛽𝛽𝛽𝛽 cos 𝜃𝜃) 𝐸𝐸𝑖𝑖 = 2 cos 𝜃𝜃𝑒𝑒 −𝑖𝑖𝑖𝑖𝑖𝑖 sin 𝜃𝜃 sin (𝛽𝛽𝛽𝛽 cos 𝜃𝜃) 𝜂𝜂 13