Maxwell's Equations and Electromagnetic Waves in Physics for Scientists & Engineers 2, Study notes of Geology

Download Maxwell's Equations and Electromagnetic Waves in Physics for Scientists & Engineers 2 and more Study notes Geology in PDF only on Docsity! March 21, 2005 Physics for Scientists&Engineers 2 1 Physics for Scientists & Engineers 2 Spring Semester 2005 Lecture 32 March 21, 2005 Physics for Scientists&Engineers 2 2 Review ! Maxwell’s Equations Name Equation Description Gauss’ Law for Electric Fields ! E • d ! A = qenc !0 " Relates the net electric flux to the net enclosed electric charge Gauss’ Law for Magnetic Fields ! B • d ! A" = 0 States that the net magnetic flux is zero (no magnetic charge) Faraday’s Law ! E • d ! s" = # d$B dt Relates the induced electric field to the changing magnetic flux Ampere-Maxwell Law ! B • d ! s" = µ0!0 d$E dt + µ0ienc Relates the induced magnetic field to the changing electric flux and to the current March 21, 2005 Physics for Scientists&Engineers 2 3 Review (2) ! We make the Ansatz that the magnitude of the electric and magnetic fields in electromagnetic waves are given by the form ! where k = 2!/" is the angular wave number and # = 2!f is the angular frequency of a wave with wavelength " and frequency f ! We assume that the electric field is in the y direction and the magnetic field is in the z direction ! Now we want to show that these equations satisfy Maxwell’s Equations E( ! r ,t) = E max sin kx !"t( ) B( ! r ,t) = B max sin kx !"t( ) ! E( ! r ,t) = E( ! r ,t) ! e y ! B( ! r ,t) = B( ! r ,t) ! e z March 21, 2005 Physics for Scientists&Engineers 2 4 Gauss’ Law for Electric Fields ! Let’s start with Gauss’ Law for electric fields ! For an electromagnetic wave, there is no enclosed charge (qenc = 0), so we must show that our solution satisfies ! We can draw a rectangular Gaussian surface around a snapshot of the wave as shown to the right ! For the faces in y-z and x-y planes ! The faces in the x-z planes will contribute +EA and -EA ! Thus the integral is zero and Gauss’ Law for electric fields is satisfied ! E • d ! A = 0"! ! E ! d ! A " ! E #d ! A = 0 March 21, 2005 Physics for Scientists&Engineers 2 5 Gauss’ Law for Magnetic Fields ! For Gauss’ Law for magnetic fields we must show ! We use the same surface that we used for the electric field ! For the faces in the y-z and x-z planes ! The faces in the x-y planes will contribute +BA and -BA ! Thus our integral is zero and Gauss’ Law for magnetic fields is satisfied ! B • d ! A"! = 0 ! B ! d ! A " ! B • d ! A = 0 March 21, 2005 Physics for Scientists&Engineers 2 6 Faraday’s Law ! Now let’s tackle Faraday’s Law ! To evaluate the integral on the left side, we assume an integration loop in the x-y plane that has a width dx and height h as shown by the gray box in the figure ! The differential area of this rectangle is ! The electric and magnetic fields change as we move in the x direction ! E • d ! s"! = " d# B dt d ! A = ! ndA = ! nhdx ! n in + z direction( ) ! E(x) ! ! E(x + dx) = ! E(x)+ d ! E March 21, 2005 Physics for Scientists&Engineers 2 7 Faraday’s Law (2) ! The integral around the loop is ! We can split this integral over a closed loop into four pieces, integrating from • a to b • b to c • c to d • d to a ! The contributions to the integral parallel to the x-axis, integrating from b to c and from d to a, are zero because the electric field is always perpendicular to the integration direction ! For the integrations along the y-direction, a to b and c to d, one has the electric field parallel to the direction of the integration • The scalar product simply reduces to a conventional product ! E • d ! s"! = E + dE( )h " Eh = dE #h March 21, 2005 Physics for Scientists&Engineers 2 8 Faraday’s Law (3) ! Because the electric field is independent of the y-coordinate, it can be taken out of the integration • The integral along each of the segments in the ±y direction is a simple product of the integrand, the electric field at the corresponding y-coordinate, times the length of the integration interval, h ! The right hand side is given by ! So we have ! d" B dt = !A dB dt = !hdx dB dt hdE = !hdx dB dt " dE dx = ! dB dt