March 21, 2005 Physics for Scientists&Engineers 2 1Physics for Scientists &Physics for Scientists &EngineersEngineers 22Spring Semester 2005Lecture 32March 21, 2005 Physics for Scientists&Engineers 2 2ReviewReview! Maxwell’s EquationsName 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$Bdt Relates the induced electric field to the changing magnetic flux Ampere-Maxwell Law !B • d!s""=µ0!0d$Edt+µ0ienc Relates the induced magnetic field to the changing electric flux and to the current March 21, 2005 Physics for Scientists&Engineers 2 3Review (2)Review (2)! We make the Ansatz that the magnitude of the electric and magnetic fields inelectromagnetic waves are given by the form! where k = 2!/" is the angular wave number and # = 2!f is the angular frequencyof a wave with wavelength " and frequency f! We assume that the electric field is in the y direction and the magnetic field isin the z direction! Now we want to show that these equations satisfy Maxwell’s Equations E(!r ,t) = Emaxsin kx !"t( )B(!r ,t ) = Bmaxsin kx !"t( ) !E(!r,t) = E(!r,t)!ey!B(!r,t) = B(!r,t)!ezMarch 21, 2005 Physics for Scientists&Engineers 2 4GaussGauss’’ Law for Electric Fields Law for Electric Fields! Let’s start with Gauss’ Law for electric fields! For an electromagnetic wave, there is no enclosed charge (qenc = 0), sowe 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 = 0March 21, 2005 Physics for Scientists&Engineers 2 5GaussGauss’’ Law for Magnetic Fields 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 fieldsis satisfied !B • d!A"!= 0 !B ! d!A " !B • d!A = 0March 21, 2005 Physics for Scientists&Engineers 2 6FaradayFaraday’’s Laws Law! Now let’s tackle Faraday’s Law! To evaluate the integral on the left side,we assume an integration loop in the x-yplane that has a width dx and height h asshown 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#Bdt d!A =!ndA =!nhdx !n in + z direction( ) !E(x) ! !E(x + dx) =!E(x) + d!EMarch 21, 2005 Physics for Scientists&Engineers 2 7FaradayFaraday’’s Law (2)s Law (2)! The integral around the loop is! We can split this integral over a closed loopinto 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, integratingfrom b to c and from d to a, are zero because the electric field isalways perpendicular to the integration direction! For the integrations along the y-direction, a to b and c to d, one has theelectric 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 # hMarch 21, 2005 Physics for Scientists&Engineers 2 8FaradayFaraday’’s Law (3)s Law (3)! Because the electric field is independentof the y-coordinate, it can be taken outof the integration• The integral along each of the segments inthe ±y direction is a simple product of theintegrand, the electric field at thecorresponding y-coordinate, times the lengthof the integration interval, h! The right hand side is given by! So we have!d"Bdt= ! AdBdt= !hdxdBdthdE = !hdxdBdt " dEdx= !dBdtMarch 21, 2005 Physics for Scientists&Engineers 2 9FaradayFaraday’’s Law (4)s Law (4)! The derivatives dE/dx and dB/dt are taken with respect to a singlevariable, although both E and B depend on both x and t! Thus we can more appropriately write! Taking our assumed form for the electric and magnetic fields we canexecute the derivatives!E!x= "!B!t!E!x=!!xEmaxsin kx "#t( )( )= kEmaxcos kx "#t( )!B!t=!!tBmaxsin kx "#t( )( )= "#Bmaxcos kx "#t( )March 21, 2005 Physics for Scientists&Engineers 2 10FaradayFaraday’’s Law (5)s Law (5)! Which gives us! We can relate the angular frequency # and the angular wave number k as! We can then write! Which implies that our assumed solution satisfies Faraday’s Law as long asE/B = ckEmaxcos kx !"t( )= ! !"Bmaxcos kx !"t( )( )!k=2"f2"#$%&'()= f#= c c is the speed of light( )EmaxBmax=!k= c " EB= cMarch 21, 2005 Physics for Scientists&Engineers 2 11Ampere-Maxwell LawAmpere-Maxwell Law! For electromagnetic waves there is no current! To evaluate the integral on the left hand sideof this equation, we assume an integration loopin the plane that has a width I and height hdepicted by the gray box in the figure! The differential area of this rectangle is oriented along the +y direction! The integral around the loop in a counter-clockwise direction(a to b to c to d to a) is given by! The parts of the loop that are parallel to the axis do not contribute !B • d!s"!=µ0"0d#Edt !B • d!s"!= " B + dB( )h + Bh = "dB #hMarch 21, 2005 Physics for Scientists&Engineers 2 12Ampere-Maxwell Law (2)Ampere-Maxwell Law (2)µ0!0d"Edt=µ0!0dE # Adt=µ0!0dE # h # dxdt!dB " h =µ0#0dE " h " dxdt! The right hand side can be written as! Substituting back into the Ampere-Maxwell relation we get! Simplifying and expressing this equation in terms of partial derivativesas we did before we get! Putting in our assumed solutions gives us!"B"x=µ0#0"E"t! kBmaxcos kx !"t( )( )= !µ0#0"Emaxcos kx !"t( )March 21, 2005 Physics for Scientists&Engineers 2 13Ampere-Maxwell Law (3)Ampere-Maxwell Law (3)! We can rewrite the previous equation as! This relationship also holds for the electric and magnetic field at anytime! So our assumed solutions satisfy the Ampere Maxwell law if the ratioof E/B isEmaxBmax=kµ0!0"=1µ0!0cEB=1µ0!0c1µ0!0cMarch 21, 2005 Physics for Scientists&Engineers 2 14The Speed of LightThe Speed of Light! Our solutions for Maxwell’s Equations were correct if! We can see that! Thus the speed of an electromagnetic wave can be
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