Physics for Scientists & Engineers 2

Transcription

1 Review Maxwell s Equations Physis for Sientists & Engineers 2 Spring Semester 2005 Leture 32 Name Equation Desription Gauss Law for Eletri E d A = q en Fields " 0 Gauss Law for Magneti Fields Faraday s Law Ampere-Maxwell Law Relates the net eletri flux to the net enlosed eletri harge " B d A = 0 States that the net magneti " E d s = # d$ B d$ " B d s = E 0 + i en flux is zero (no magneti harge) Relates the indued eletri field to the hanging magneti flux Relates the indued magneti field to the hanging eletri flux and to the urrent Marh 2, 2005 Physis for Sientists&Engineers 2 Marh 2, 2005 Physis for Sientists&Engineers 2 2 Review (2) We make the Ansatz that the magnitude of the eletri and magneti fields in eletromagneti waves are given by the form E( r,t) = E max sin kx "t B( r,t) = B max sin kx "t where k = 2/" is the angular wave number and # = 2f is the angular frequeny of a wave with wavelength " and frequeny f We assume that the eletri field is in the y diretion and the magneti field is in the z diretion E( r,t) = E( r,t) e y B( r,t) = B( r,t) e z Now we want to show that these equations satisfy Maxwell s Equations Gauss Law for Eletri Fields Let s start with Gauss Law for eletri fields For an eletromagneti wave, there is no enlosed harge (q en = 0), so we must show that our solution satisfies " E d A = 0 We an draw a retangular Gaussian surfae around a snapshot of the wave as shown to the right For the faes in y-z and x-y planes E d A " E #d A = 0 The faes in the x-z planes will ontribute +EA and -EA Thus the integral is zero and Gauss Law for eletri fields is satisfied Marh 2, 2005 Physis for Sientists&Engineers 2 3 Marh 2, 2005 Physis for Sientists&Engineers 2 4

2 Gauss Law for Magneti Fields For Gauss Law for magneti fields we must show " B d A = 0 We use the same surfae that we used for the eletri field For the faes in the y-z and x-z planes B d A " B d A = 0 The faes in the x-y planes will ontribute +BA and -BA Thus our integral is zero and Gauss Law for magneti fields is satisfied Faraday s s Law Now let s takle Faraday s Law " E d s = " d# B To evaluate the integral on the left side, we assume an integration loop in the x-y plane that has a wih dx and height h as shown by the gray box in the figure The differential area of this retangle is d A = nda = nhdx n in + z diretion The eletri and magneti fields hange as we move in the x diretion E(x) E(x + dx) = E(x) + d E Marh 2, 2005 Physis for Sientists&Engineers 2 5 Marh 2, 2005 Physis for Sientists&Engineers 2 6 Faraday s s Law (2) The integral around the loop is E d s " = E + de h " Eh = de #h We an split this integral over a losed loop into four piees, integrating from a to b b to to d d to a The ontributions to the integral parallel to the x-axis, integrating from b to and from d to a, are zero beause the eletri field is always perpendiular to the integration diretion For the integrations along the y-diretion, a to b and to d, one has the eletri field parallel to the diretion of the integration The salar produt simply redues to a onventional produt Faraday s s Law (3) Beause the eletri field is independent of the y-oordinate, it an be taken out of the integration The integral along eah of the segments in the ±y diretion is a simple produt of the integrand, the eletri field at the orresponding y-oordinate, times the length of the integration interval, h The right hand side is given by d" B = A db db = hdx So we have hde = hdx db " de dx = db Marh 2, 2005 Physis for Sientists&Engineers 2 7 Marh 2, 2005 Physis for Sientists&Engineers 2 8

3 Faraday s s Law (4) The derivatives de/dx and db/ are taken with respet to a single variable, although both E and B depend on both x and t Thus we an more appropriately write E x = " B t Taking our assumed form for the eletri and magneti fields we an exeute the derivatives E x = x E max sin kx " #t B t = t ( ) = ke max os( kx " #t) ( B max sin( kx " #t)) = "# B max os( kx " #t) Whih gives us ke max os kx "t Faraday s s Law (5) We an relate the angular frequeny # and the angular wave number k as k = 2" f $ 2" ' % & # ( ) We an then write = (" B max os( kx "t)) = f # = ( is the speed of light) E max = B max k = " E B = Whih implies that our assumed solution satisfies Faraday s Law as long as E/B = Marh 2, 2005 Physis for Sientists&Engineers 2 9 Marh 2, 2005 Physis for Sientists&Engineers 2 0 Ampere-Maxwell Law For eletromagneti waves there is no urrent " B d s d# = " E 0 To evaluate the integral on the left hand side of this equation, we assume an integration loop in the plane that has a wih I and height h depited by the gray box in the figure The differential area of this retangle is oriented along the +y diretion The integral around the loop in a ounter-lokwise diretion (a to b to to d to a) is given by B d s " = " B + db h + Bh = "db # h The parts of the loop that are parallel to the axis do not ontribute Ampere-Maxwell Law (2) The right hand side an be written as 0 d" E = 0 de # A = 0 de #h # dx Substituting bak into the Ampere-Maxwell relation we get de " h "dx db "h = # 0 Simplifying and expressing this equation in terms of partial derivatives as we did before we get "B "x = µ "E 0# 0 "t Putting in our assumed solutions gives us ( kb max os( kx "t)) = # 0 "E max os( kx "t) Marh 2, 2005 Physis for Sientists&Engineers 2 Marh 2, 2005 Physis for Sientists&Engineers 2 2

4 Ampere-Maxwell Law (3) We an rewrite the previous equation as E max k = B max 0 " = 0 This relationship also holds for the eletri and magneti field at any time E B = 0 So our assumed solutions satisfy the Ampere Maxwell law if the ratio of E/B is 0 Marh 2, 2005 Physis for Sientists&Engineers 2 3 The Speed of Light Our solutions for Maxwell s Equations were orret if E B = and E B = 0 We an see that = 0 " = 0 Thus the speed of an eletromagneti wave an be expressed in terms of two fundamental onstants related to eletri fields and magneti fields, the magneti permeability and the eletri permittivity of the vauum If we put in the values of these onstants that we have been using we get =.26 0 "6 H/m ( "2 F/m) = m/s Marh 2, 2005 Physis for Sientists&Engineers 2 4 The Speed of Light (2) This similarity implies that all eletromagneti waves travel at the speed of light Further this result implies that light is an eletromagneti wave The speed of light plays an important role in the theory of relativity The speed of light is always the same in any referene frame Thus if you send an eletromagneti wave out in a speifi diretion, any observer, regardless of whether that observer is moving toward you or away from you, will see that wave moving at the speed of ligh This amazing result leads to the theory of relativity The speed of light an be measured extremely preisely, muh more preisely than we an determine the meter from the original referene standard. So now the speed of light is defined as preisely = 299,792,458 m/s The Eletromagneti Spetrum All eletromagneti waves travel at the speed of light However, the wavelength and frequeny of eletromagneti waves an vary dramatially The speed of light, the wavelength ", and the frequeny f are related by = f Examples of eletromagneti waves inlude light, radio waves, mirowaves, x-rays, and gamma rays This diverse spetrum is illustrated on the next slide Marh 2, 2005 Physis for Sientists&Engineers 2 5 Marh 2, 2005 Physis for Sientists&Engineers 2 6

5 The Eletromagneti Spetrum (2) Eletromagneti Spetrum (3) Eletromagneti waves exist with wavelengths ranging from 000 m to less than 0-2 m and frequenies ranging from 0 6 to 0 2 Hz Certain ranges of wavelength and frequeny have names that identify the most ommon appliation of those eletromagneti waves Visible light refers to eletromagneti waves ranging in wavelength from 400 nm to 700 nm The response of the human eye is peaked around 550 nm (green) and drops off quikly away from that wavelength Other wavelengths of eletromagneti waves are invisible to the human eye However, we an still detet the eletromagneti waves by other means. For example, we an feel eletromagneti waves in the infrared (wavelengths just longer than visible up to around m) as warmth. Marh 2, 2005 Physis for Sientists&Engineers 2 7 Marh 2, 2005 Physis for Sientists&Engineers 2 8