The behavior of electric and magnetic waves can be fully described by a setof four equations (which we have learned already). BFaraday’s Law of induction E t DAmpere’s Law H J tGauss’s Law for electricity D vGauss’s Law for magnetism B 0
And the constitutive relations: D E B H J EThey relate the electromagnetic field to the properties of the material, inwhich the field exists. Together with the Maxwell’s 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.
Integral form BFaraday’s Law of induction E dl t ds L s DAmpere’s Law H d l t J d s L sGauss’s Law for electricity D ds v dv S vGauss’s Law for magnetism B d s 0 S
Example 6.1: In a conductive material we may assume that the conductivecurrent density is much greater than the displacement current density. Showthat the Maxwell’s equations can be put in a form of a Diffusion equation in thismaterial. B We 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 and B 0 t Is the diffusion equation with a diffusion coefficient D = 1/( 0)
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: 2 d Bx ( z ) 2 j 0 Bx ( 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 0 j j 0
The magnitude of the magnetic flux density decays exponentially in the zdirection from the surface into the conductor z Bx ( z) B0e where 7 7 1 f 0 60 4 10 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.
Example 6.3: Derive the equation of continuity starting from the Maxwell’s equationsThe Gauss’s law: D v v DTaking time derivatives: D t t t DFrom the Ampere’s law H J t vTherefore: H J t vThe equation of continuity: J t
It is frequently needed to determine the direction the power is flowing. ThePoynting’s Theorem is the tool for such tasks.We consider an arbitraryshaped volume:Recall: B E t D H J tWe take the scalar product of E and subtract it from the scalar product of H. B D H E E H H E J t t
Using the vector identity ( A B) B A A B Therefore: B D ( E H) H E E J t tApplying the constitutive relations to the terms involving time derivatives, we get: B D 1 1 2 2 H E H H E E H E t t 2 t t 2Combining (6.9.2) and (6.9.3) and integrating both sides over the same v…
1 2 2 ( E H ) dv H E dv E Jdv v t v 2 vApplication of divergence theorem and the Ohm’s law lead to the PT: 1 2 2 2 (E H ) ds H E dv E dv s t v 2 vHere S E His the Poynting vector – the power density and thedirection of the radiated EM fields in W/m2.
The Poynting’s 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 is 1 2 2 w H E 2Power loss density is 2 pL EThe differential form of the Poynting’s Theorem: w S pL t
Example 6.4: Using the Poynting’sTheorem, 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 theresistor’s radius is much less than its length. The magnitude of the electric field is E V0 L and it is in the direction of the current. The magnitude of the magnetic field intensity at the outer surface of the resistor: H I 2 a
The Poynting’s vector S E His 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. I J 2 a V0 I d I V0 2The PT: 2 aL (0 0) dv 2 a L L 2 a dt v a L V0 I V0 IThe electromagnetic energy of a battery is completely absorbed withthe resistor in form of heat.
Example 6.5: Using Poynting’sTheorem, 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 E 2 W a b 2The total magnetic energy stored in the capacitor is zero.
The time derivative of the electric energy is dW 2 dE a bE dt dtThis 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 dE 2 aH a H dt 2 dt
The power flow would be: PS E H ds sIn our situation: ds 2 ab u rand S u r 1 2 dETherefore: PS 2 abE H a bE dt dWWe observe that PS dtThe energy is conserved in the circuit.
Frequently, a temporal variation of EM fields is harmonic;therefore, we may use a phasor representation: j t E ( x, y, z, t ) R e E ( x, y , z )e j t H ( 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).Maxwell’s Eqn in E (r ) j H (r )phasor form: H (r ) j E (r ) J (r ) E ( r ) v (r ) B ( r ) 0
Power is a real quantity and, keeping in mind that: j t j t j t j t R e E ( r )e R e H ( r )e R e E (r )e H ( r )e complex conjugate * A A Since Re A 2Therefore: * * 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
Therefore, the Poynting’s theorem in phasors is: * 2 2 2 E (r ) H ( r ) ds j H E dv E dv s v v Total power radiated The energy stored The power dissipated from the volume within the volume within the volume Indicates that the power (energy) is reactive
Example 6.6: Compute the frequency at which the conduction current equals thedisplacement current in copper.Using the Ampere’s law in the phasor form, we write: H (r ) J (r ) j E (r )Since J Eand J (r ) J d (r ) E (r ) j E (r )Therefore: 7 5.8 10 18Finally: f 1 1.04 10 Hz 2 2 9 0 2 10 36At much higher frequencies, cooper (a good conductor) acts like a dielectric.
Example 6.7: The fields in a free space are: 4 z uz E E 10 cos t ux; H 3 120Determine the Poynting vector if the frequency is 500 MHz.In a phasor notation: 4 z 4 z j 3 10 j 3 E (r ) 10 e ux H (r ) e uy 120And the Poynting vector is: 2 1 * 10 S av ( r ) R e E (r ) H (r ) uz 0.133 u z 2 2 120 HW 5 is ready
The diffusion equation is a partial differential equation which describes density fluctuations in a material undergoing diffusion. Diffusion is the movement of particles of a substance from an area of high concentration to an area of low concentration, resulting in the uniform distribution of the substance. 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
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