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The Maxwell Equations Farady’s law and the Electromagnetic “force” (EMF). Faraday's law relates the electromotive force, , with the change of the magnetic flux with time, 1 d c dt The electromagnetic force around the circuit through which the flux is passed is E d ˆ da E n C S We note that 1 B 1 d 1d 1 B d x d B d v nˆda c dt c dt S c S c S t Hence when the loop is not moving, we have 1 B E 0 c t The Maxwell equations in vacuum E 4 B 0 1 B 1 E 4 E 0 B J c t c t c 1 Implicit in those are the continuity equation, J 0 t and the Lorentz force vB F q E c The Maxwell equations in macroscopic media D 4 f B 0 1 B 1 D 4 E 0 H Jf c t c t c with D E 4P H B 4M Boundary conditions Let the fields in two neighboring media be denoted by 1 and 2. Let the surface charge density on the surface between the two media be boundary conditions are ( D2 D1 ) nˆ 4 ( E2 E1 ) nˆ 0 and let the surface current density there be K . Then the ( B2 B1 ) nˆ 0 4 nˆ ( H 2 H1 ) K c 2 Scalar and vector potentials A Let us introduce the vector potential such that Then, from the Maxwell’s equations, we have B A . This ensures that B 0 . 1 A 0 E c t and therefore we choose 1 A E c t These choices satisfy the two homogeneous Maxwell’s equations. The other two equations determine the scalar and the vector potentials, 1 2 A 4 c t 2 1 A 1 4 2 A 2 2 ( A ) J c t c t c We note that both the electric and the magnetic field are unchanged by the transformation A A' A 1 ' c t and therefore we can choose the gauge 3 1 A 0 c t to obtain the Maxwell’s equations in the form 1 2 2 2 4 c t 1 2 A 4 (Lorentz gauge) 2 A 2 2 J c t c But of course, we could have stayed with A 0 , and then to obtain 2 2 4 1 2 A 1 4 2 A 2 2 ( ) J c t c t c (Coulomb gauge) Poynting theorem The rate of work done by the electromagnetic fields on a moving charge q is qv E . For a system of current density, that rate will be 1 D d r J E d r c E ( H ) E 4 t Using ( E H ) H ( E ) E ( H ) we find 4 1 D B dr J E 4 dr c ( E H ) E t H t Denoting the total energy density of the electromagnetic field by U, 1 U ED BH 8 we obtain the conservation law U S J E t total energy rate mechanical work with the Poynting vector c S EH 4 Exercise: A segment of a wire of length L and radius a, carrying the current I. A voltage drop V is applied on the wire. Find the Poynting vector. Solution: We us cylindrical coordinates, placing the wire along the z-direction. The magnetic field produced by the current in the wire (on the surface of the wire) is 2I B ̂ ca 5 and the electric field is V E zˆ , L The Poynting vector at the surface of the wire is c I V S EH ˆ 4 2a L , which is the power supplied by the battery, 2aL . P IV , divided by the surface area of the wire, Electromagnetic waves In the absence of sources (i.e., no free charges or free currents) the Maxwell’s equations read D 0 B 0 1 B 1 D E 0 H 0 c t c t Let us describe the medium by its dielectric constant, D E , and magnetic permeability, B H . We then have 6 E 0 B 0 1 B E E 0 B 0 c t c t We see that each of the filed components satisfy the wave equation 1 2u u 2 2 0 v t 2 with the velocity v c The solution is (in general) ik r it ik r it u(r , t ) Ae Be with the dispersion relation relating the wave-vector with the frequency k v c Let us now solve for the electromagnetic fields. We write iknˆ r it E (r , t ) E0e iknˆ r it B(r , t ) B0e 7 with the convention that the physical fields are the real parts of the above expressions. From the homogeneous Maxwell’s equations we have that E0 nˆ 0 B0 nˆ 0 and therefore both the electric and the magnetic fields are perpendicular to the direction of the wave propagation. Such a wave is called "transverse wave". From the other two Maxwell’s equations we have B0 nˆ E0 The Poynting vector, averaged over a period, is given by the real part of c * S EH 8 Explanation: c 1 * 1 * c * * * * S ( E E ) ( H H ) E H E H E H E H 4 2 2 16 c * * E H E H 16 In our case c S 8 2 | E0 | nˆ The energy density, averaged over a period is 8 1 u 16 * 1 * E E B B and in our case u 2 | E0 | 8 Electromagnetic waves in a conducting medium ( We consider the charge-free case.) The Maxwell equations are D E 0 B 0 1 B 1 D 4 E 0 H Jf c t c t c Using Ohm’s law, relating the current density to the electric field via the conductivity J E the fourth equation becomes 1 E 4 B E c t c Therefore we now find 9 2 E 4 E E 2 2 0 2 2 c t B c t B B Using again a plane-wave solution, the dispersion law (relating the wave vector to the frequency) is now k 2 2 4 1 i c2 and the wave is damped. Approximately, we find 2 c c 2 k (1 i ) c k i 4 1 4 1 In a good conductor (high conductivity, law frequency), the electromagnetic wave penetrates only over a length given by c 2 which is called "skin-depth". Exercise: A wave of frequency , polarized in the x-direction, propagates along the z-direction in a good conductor which occupies the z>0 space. Find the current in the conductor. 10 Solution: Let us that , 1. The wave is then given by E E0 xˆeit e z (1i ) , and the current density established in the conductor is J E0 xˆeit e z (1i ) / Let us consider a slab of dimensions direction) is h dz . The current flowing in the slab (along the x- dI E0e it e z (1 i ) / hdz and the total current becomes I E0eit e z (1i ) / hdz E0 0 h 1 i eit Taking the real part to obtain the `true’ current, we find I E0h cos(t 4 ) 2 We see that the current is not in-phase with the field. 11 Waveguides Consider a rectangular waveguide of cross section of dimensions a and b and axis along the zdirection. In the absence of charge and current distributions, the electric and magnetic fields satisfy 2 1 E 2 E 2 2 0 H c t H When the wave propagates along the z-direction, and the electric field is only in the x-y plane, then E xˆEx yˆE y H xˆH x yˆH y zˆH z and all the z-dependence of the fields is in the form ik g z e . Denoting ck we have 2H z 2H z (k 2 k g2 ) H z 0 2 2 x y 1 B 0 we then find From the Maxwell’s equation E c t k k Ex H y , E y H x kg kg 12 and from his other equation ikEx 1 D H 0 c t we have H z H z ik g H y 0, ikEy ik g H x 0 y x Therefore, k 2 H z ik g 1 H x k x g k 2 H z ik g 1 H y k y g and it is sufficient to find Hz to find all fields. Now we solve for that component by variables separation. Writing H z ( x, y) h1 ( x)h2 ( y) we have d 2 h1 2 k h1 0, x 2 dx d 2h2 2 k h2 0 y 2 dy with k x2 k y2 k 2 k g2 0 13 We now have the following solution Hz e ikg z E x i Ey i ( A1 sin k x x sin k y y A2 sin k x x cos k y y A3 cos k x x sin k y y A4 cos k x x cos k y y) kk y k 2 g (1 [ k 2 1 ikg z ] ) e ( A1 sin k x x cos k y y A2 sin k x x sin k y y A3 cos k x x cos k y y A4 cos k x x sin k y y) kg kkx k ik z (1 [ ]2 ) 1 e g ( A1 cos k x x sin k y y A2 cos k x x cos k y y A3 sin k x x sin k y y A4 sin k x x cos k y y) 2 kg kg Now we need to apply boundary conditions. Taking the surface of the waveguide as a conductor then Ey 0 at x=0 and x=a gives A1 A2 0 and k x m / a where m is an integer. Taking Ex 0 at y=0 and y=b gives A1 A3 0 and k y n / b where n is an integer. Therefore the final solution is Hz e ik g z Amn cos( nm mx ny ) cos( ) a b with 14 m n k k a b 2 2 g 2 2 limiting the allowed values of m and n! 15