1.The behavior of electric and magnetic waves can be fully described by a setof four equations (which we have learned already). BFaradays Law of inductionE t DAmperes LawHJ tGausss Law for electricity D vGausss Law for magnetism B 0

2. And the constitutive relations: DEB H JEThey relate the electromagnetic field to the properties of the material, inwhich the field exists. Together with the Maxwells equations, theconstitutive relations completely describe the electromagnetic field.Even the EM fields in a nonlinear media can be described through anonlinearity existing in the constitutive relations. 3. Integral formBFaradays Law of induction E dl tds LsDAmperes Law H d ltJ d sLsGausss Law for electricity D ds vdv SvGausss Law for magnetism B d s 0 S 4. Example 6.1: In a conductive material we may assume that the conductivecurrent density is much greater than the displacement current density. Showthat the Maxwells equations can be put in a form of a Diffusion equation in thismaterial.BWe can write:E t and, neglecting the H J E displacement current: Taking curl of (6.5.2): H E B 2 B B Expanding the LHS: 0 0 t 2 B The first term is zero andB 0 t Is the diffusion equation with a diffusion coefficient D = 1/(0) 5. Example 6.2: Solve the diffusion equation for the case of the magnetic fluxdensity Bx(z,t) near a planar vacuum-copper interface, assuming for copper: = 0 and = 5.8 x 107 S/m. Assume that a 60-Hz time-harmonic EM signal isapplied. Assuming ej t time-variation, the diffusion equation is transformed to the ordinary differential equation: 2d Bx ( z )2j 0Bx ( z ) dzWhere z is the normal coordinate to the boundary. Assuming a variation inthe z-direction to be Bx(z) = B0e- z, we write:2 j 0jj 0 6. The magnitude of the magnetic flux density decays exponentially in the zdirection from the surface into the conductor z Bx ( z) B0e where 77 1 f 060 410 5.8 10 117.2 mThe quantity = 1/ is called a skin depth - thedistance over which the current (or field) falls to 1/e ofits original value. For copper,= 8.5 mm. 7. Example 6.3: Derive the equation of continuity starting from the Maxwells equationsThe Gausss law:D vv DTaking time derivatives:D t ttDFrom the Amperes lawH Jt vTherefore:H J tvThe equation of continuity:Jt 8. It is frequently needed to determine the direction the power is flowing. ThePoyntings Theorem is the tool for such tasks.We consider an arbitraryshaped volume:Recall:BEtDH JtWe take the scalar product of E and subtract it from the scalar product of H.B D H E E H HE J tt 9. Using the vector identity( A B) B AABTherefore: BD( E H) H E E J ttApplying the constitutive relations to the terms involving time derivatives, we get:B D1 1 2 2H E H H E EH Et t2t t 2Combining (6.9.2) and (6.9.3) and integrating both sides over the same v 10. 1 2 2 ( E H ) dvH E dvE Jdvv t v2vApplication of divergence theorem and the Ohms law lead to the PT: 12 2 2 (EH ) ds H E dvE dv s t v 2 vHere SE His the Poynting vector the power density and thedirection of the radiated EM fields in W/m2. 11. The Poyntings Theorem states that the power that leaves a region isequal to the temporal decay in the energy that is stored within thevolume minus the power that is dissipated as heat within it energyconservation.EM energy density is12 2 w H E2Power loss density is 2pL EThe differential form of the Poyntings Theorem: w SpL t 12. Example 6.4: Using the PoyntingsTheorem, calculate the power that is dissipatedin the resistor as heat. Neglect the magneticfield that is confined within the resistor andcalculate its value only at the surface. Assumethat the conducting surfaces at the top and thebottom of the resistor are equipotential and theresistors radius is much less than its length.The magnitude of the electric field isE V0 Land it is in the direction of the current.The magnitude of the magnetic field intensity at the outer surface of the resistor:H I2 a 13. The Poyntings vector S EHis into the resistor. There is NO energy stored in theresistor. The magnitude of the current density is in thedirection of a current and, therefore, the electric field.IJ 2aV0I dI V02The PT: 2 aL(0 0) dv 2a L L 2 adt v a L V0 IV0 IThe electromagnetic energy of a battery is completely absorbed withthe resistor in form of heat. 14. Example 6.5: Using PoyntingsTheorem, calculate the power that isflowing through the surface area at theradial edge of a capacitor. Neglect theohmic losses in the wires, assume thatthe radius of the plates is much greaterthan the separation between them: a >>b.Assuming the electric field E is uniform and confined between the plates, the totalelectric energy stored in the capacitor is: 2 E2 W a b 2The total magnetic energy stored in the capacitor is zero. 15. The time derivative of the electric energy isdW 2 dE a bE dtdtThis is the only nonzero term on the RHS of PT since an ideal capacitor does notdissipate energy.We express next the time-varying magnetic field intensity in terms of thedisplacement current. Since no conduction current exists in an ideal capacitor: E H dl t ds sTherefore:dE 2 a dE2 aH aHdt 2 dt 16. The power flow would be: PSE H ds sIn our situation: ds2 ab u rand S u r12 dETherefore:PS 2 abE H a bEdtdWWe observe thatPS dtThe energy is conserved in the circuit. 17. Frequently, a temporal variation of EM fields is harmonic;therefore, we may use a phasor representation:j tE ( x, y, z, t )R e E ( x, y , z )ej tH ( x, y, z, t ) R e H ( x, y , z )eIt may be a phase angle between the electric and the magnetic fieldsincorporated into E(x,y,z) and H(x,y,z).Maxwells Eqn inE (r )jH (r )phasor form:H (r ) jE (r )J (r )E ( r )v(r ) B ( r )0 18. Power is a real quantity and, keeping in mind that:j t j tj tj tR e E ( r )e R e H ( r )e R e E (r )eH ( r )e complex conjugate*A A Since Re A2Therefore:* * E (r ) E (r ) H (r ) H (r ) R e E (r )R e H (r )2 2**** E (r ) H (r ) E (r )H (r ) E (r )H (r ) E (r ) H (r )4 Taking the time average, we obtain the average power as: 1 *S av ( r ) R e E (r ) H (r ) 2 19. Therefore, the Poyntings theorem in phasors is: * 222E (r )H ( r ) dsjHE dvE dvsv v Total power radiated The energy stored The power dissipated from the volumewithin the volume within the volumeIndicates that the power (energy) is reactive 20. Example 6.6: Compute the frequency at which the conduction current equals thedisplacement current in copper.Using the Amperes law in the phasor form, we write:H (r )J (r ) j E (r )Since JEand J (r ) J d (r ) E (r ) j E (r )Therefore: 75.8 10 18Finally: f1 1.04 10Hz2 29 0210 36At much higher frequencies, cooper (a good conductor) acts like a dielectric. 21. Example 6.7: The fields in a free space are: 4 z uzE E10 cos tux; H3120Determine the Poynting vector if the frequency is 500 MHz.In a phasor notation: 4 z4 z j 3 10 j 3 E (r ) 10 e ux H (r ) euy120And the Poynting vector is: 21 * 10 S av ( r ) R e E (r ) H (r )uz0.133 u z 2 2 120 HW 5 is ready 22. The diffusion equation is a partial differential equation whichdescribes density fluctuations in a material undergoing diffusion.Diffusion is the movement ofparticles of a substance from anarea of high concentration to anarea of low concentration, resultingin the uniform distribution of thesubstance. Similarly, a flow of free charges in a material, where a charge difference between two locations exists, can be described by the diffusion equation.Back 23. THE ENDReference Journal of Scientific Exploration