EEE340Lecture 341 The time domain fields are: Physically, the H field is perpendicular to the E field, and H is 45 degree lacking. By setting the decay to 36.79%, we define the skin depth. Skin depth (8-57)

EEE340Lecture 342 Example 1: Microwave ovens are operating at 2.45 GHz. The round steak has the permittivity  c =40(0.1-j0.3)  o =(4-j12)  o Find the skin depth  Solution: Is steak a good conductor?

EEE340Lecture 343 The skin depth At z= , field decays to e -1 =37% as compared to its magnitude on the surface. Example 2 Skin depth of copper at 1GHz

EEE340Lecture the dc resistance The surface (skin) resistance R s Assuming a uniform current within skin depth For a conductor wire of radius a, w = 2  a Skin-effect resistance

EEE340Lecture 345 For an aluminum wire with diameter 2.6 mm find at 10 MHz, 2 GHz (a) (b)

EEE340Lecture 346 Energy can be transported from one point to another point. (7-53a) (8-77) (7-53b) (8-78) We wish to use vector identity as follows: 8-5 Power and the Poynting Vector

EEE340Lecture 347 Using vector identity: Total power leaving the volume Rate of decrease in energy stored in electric and magnetic fields Ohmic power dissipated = (8-80) (8.82) Using divergence theorem:

EEE340Lecture 348 This is “Poyntings theorem” The Poynting vector (watts/m 2 ) instantaneous If we assume that then (8-83) (8-92)

EEE340Lecture Instantaneous versus time average power densities (Poynting vectors) The total time-average power crossing a given surface S: (8-96) (8-94)