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Electromagnetic Waves March 24, 2014 Chapter 31 1 Electromagnetic Waves ! We will see that light is an electromagnetic wave ! Electromagnetic waves have electric and magnetic fields ! We will see Maxwell’s Equations that describe electromagnetic phenomena ! We will see that the speed of light is constant and can be related to ε0 and μ0 ! We will see that electromagnetic waves can transport energy and momentum ! Electromagnetic waves can be polarized March 24, 2014 Chapter 31 2 Induced Magnetic Fields ! We have seen that a changing magnetic field induces an electric field ! Faraday’s Law of Induction tells us dΦ B ∫ Eids = − dt ! In a similar manner, a changing electric field induces a magnetic field ! Maxwell’s Law of Induction tells us dΦ E ∫ Bids = µ0ε 0 dt ! where B is the magnetic field induced in a closed loop by a changing electric flux ΦE in that loop March 24, 2014 Chapter 31 3 Circular Capacitor ! To illustrate induced magnetic fields, consider the example of a circular parallel plate capacitor ! We charge the capacitor and disconnect the battery ! The charge is constant and the electric field between the plates is constant ! There is no magnetic field ! What if the charge increases with time? ! A magnetic field is induced ! The magnitude of the induced magnetic field is the same along each line and the direction is tangential to the line March 24, 2014 Chapter 31 4 Changing Magnetic Field ! Now let’s consider a constant magnetic field ! Let’s increase the magnitude of the magnetic field while keeping it uniform in space and direction ! An electric field is induced as shown by the red loops ! The magnitude of the electric field is constant along each loop and the direction is tangential to each loop ! Note that the induced electric field points in the opposite direction from the magnetic field induced by a changing electric field March 24, 2014 Chapter 31 5 Maxwell-Ampere Law ! Ampere’s Law relates the integral around a loop of the dot product of the magnetic field and the integration direction to the current flowing through the loop ∫ Bids = µ0ienc ! Combining Maxwell’s Law of Induction and Ampere’s Law describes magnetic fields created by moving charges and by changing electric fields, the Maxwell-Ampere Law dΦ E ∫ Bids = µ0ε 0 dt + µ0ienc ! For the case of constant current, such as current flowing in a conductor, this equation reduces to Ampere’s Law ! For the case of a changing electric field without current flowing, such as the electric field between the plates of a capacitor, this equation reduces to the Maxwell Law of Induction March 24, 2014 Chapter 31 6 Displacement Current ! Looking at the Maxwell-Ampere Law dΦ E ∫ Bids = µ0ε 0 dt + µ0ienc ! The quantity ε0(dΦE/dt) must have the units of current ! This term is called the displacement current dΦ E id = ε 0 dt ! Note however that no actual current is being displaced ! We can then rewrite the Maxwell-Ampere Law as ∫ Bids = µ0 (id + ienc ) March 24, 2014 Chapter 31 7 Displacement Current ! Consider again a parallel plate capacitor with circular plates ! We place the capacitor in a circuit in which a current i is flowing while the capacitor is charging ! For a parallel plate capacitor with area A we can relate the charge q to the electric field E q = ε 0 AE ! We can get the current by taking the time derivative of the charge i= dq dE = ε0 A dt dt March 24, 2014 Chapter 31 8 Displacement Current ! Assuming that the electric field between the plates of the capacitor is uniform we can obtain an expression for the displacement current dE d ( AE ) dΦ E = ε0 A = ε0 id = ε 0 dt dt dt ! The current in the circuit is the same as the displacement current id ! Although there is no actual current flowing between the plates of the capacitor in the sense that no actual charges flow across the capacitor gap from one plate to the other, we can use the concept of displacement current to calculate the induced magnetic field March 24, 2014 Chapter 31 9 Displacement Current ! To calculate the magnetic field between the two plates of the capacitor, we assume that the volume between the plates can be replaced with a conductor of radius R carrying current id ! Thus from Chapter 28 we know that the magnetic field at a distance r from the center of the capacitor is ⎛ µ0id ⎞ B=⎜ r ⎝ 2π R 2 ⎟⎠ ! Outside the capacitor we treat the system as a current-carrying wire ! The magnetic field is µ0id B= March 24, 2014 2π r Chapter 31 10 Maxwell’s Equations ! The Maxwell-Ampere Law completes the explanation of the four equations known as Maxwell’s Equations that describe electromagnetic phenomena ! We have used these equations to describe electric fields, magnetic fields, and circuits and now electromagnetic waves Name Gauss’ Law for Electric Fields Gauss’ Law for Magnetic Fields Equation qenc ∫∫ EidA = ε 0 ∫∫ BidA = 0 Faraday’s Law dΦ B Eid s = − ∫ dt Ampere-Maxwell Law dΦ E Bid s = µ ε + µ0 ienc ∫ 0 0 dt March 24, 2014 Chapter 31 Description Relates the net electric flux to the net enclosed electric charge States that the net magnetic flux is zero (no magnetic charge) Relates the induced electric field to the changing magnetic flux Relates the induced magnetic field to the changing electric flux and to the current 12 Wave Solutions to Maxwell’s Equations ! One solution to Maxwell’s Equations is an electromagnetic wave ! Here we will assume that the electromagnetic wave propagates in vacuum (no moving charges or currents) ! We will make the Ansatz that the electric and magnetic fields in electromagnetic waves are given by the form ! ! E(r ,t) = Emax sin ( kx − ω t ) ŷ ! ! B(r ,t) = Bmax sin ( kx − ω t ) ẑ 2π k= wave number λ ω = 2π f angular frequency March 24, 2014 Chapter 31 14 Wave Solutions to Maxwell’s Equations ! The wave is traveling in the positive x-direction ! Note that there is no dependence on the y- or z-coordinates, only on the x-coordinate and time ! This type of wave is called a plane wave ! We will show that this general wave Ansatz satisfies Maxwell’s Equations March 24, 2014 Chapter 31 15 Example problem: Plane wave ! The components of the electric field in an electromagnetic wave traveling in vacuum are described by Ex = 0, Ey = 0, and Ez = 4.31 sin(kx - 1.12E8 t) V/m, where x is measured in meters and t in seconds. 1. Calculate the frequency of the wave. Answer: ω = 1.12 ×108 1s = 2π f is the angular frequency Thus the frequency is ω f= = 1.78 ×107 Hz 2π March 24, 2014 Chapter 31 16