Review Physics for Scientists Engineers 2 Spring Semester 2005 Lecture 32 Maxwell s Equations Name Gauss Law for Electric Fields Equation qenc E dA 0 Gauss Law for Magnetic Fields B dA 0 Faraday s Law E ds Ampere Maxwell Law B ds d B dt 0 0 March 21 2005 Physics for Scientists Engineers 2 1 Review 2 We make the Ansatz that the magnitude of the electric and magnetic fields in electromagnetic waves are given by the form 2 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 E dA 0 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 E r t E r t ey B r t B r t ez 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 E dA E dA 0 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 Now we want to show that these equations satisfy Maxwell s Equations March 21 2005 Physics for Scientists Engineers 2 Gauss Law for Electric Fields E r t Emax sin kx t B r t Bmax sin kx t March 21 2005 d E 0 ienc dt 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 Physics for Scientists Engineers 2 3 March 21 2005 Physics for Scientists Engineers 2 4 Gauss Law for Magnetic Fields Faraday s Law Now let s tackle Faraday s Law For Gauss Law for magnetic fields we must show B dA 0 We use the same surface that we used for the electric field B dA 0 The differential area of this rectangle is dA ndA nhdx 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 March 21 2005 Physics for Scientists Engineers 2 5 n in z direction The electric and magnetic fields change as we move in the x direction E x E x dx E x dE March 21 2005 Faraday s Law 2 Physics for Scientists Engineers 2 6 Faraday s Law 3 The integral around the loop is Because the electric field is independent of the y coordinate it can be taken out of the integration E ds E dE h Eh dE h 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 We can split this integral over a closed loop into four pieces integrating from a to b b to c c to d The right hand side is given by 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 March 21 2005 d B dt 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 For the faces in the y z and x z planes B dA E ds Physics for Scientists Engineers 2 7 d B dB dB A hdx dt dt dt So we have hdE hdx March 21 2005 dB dt dE dB dx dt Physics for Scientists Engineers 2 8 Faraday s Law 4 Faraday s Law 5 The derivatives dE dx and dB dt are taken with respect to a single variable although both E and B depend on both x and t kEmax cos kx t Bmax cos kx t Thus we can more appropriately write E B x t E Emax sin kx t kEmax cos kx t x x 9 March 21 2005 Physics for Scientists Engineers 2 10 The right hand side can be written as d B ds 0 0 dt E 0 0 To evaluate the integral on the left hand side of this equation we assume an integration loop in the plane that has a width I and height h depicted by the gray box in the figure d E dE A dE h dx 0 0 0 0 dt dt dt Substituting back into the Ampere Maxwell relation we get dB h 0 0 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 dE h dx dt Simplifying and expressing this equation in terms of partial derivatives as we did before we get B ds B dB h Bh dB h B E 0 0 x t Putting in our assumed solutions gives us kBmax cos kx t 0 0 Emax cos kx t The parts of the loop that are parallel to the axis do not contribute Physics for Scientists Engineers 2 E c B Ampere Maxwell Law 2 For electromagnetic waves there is no current March 21 2005 Which implies that our assumed solution satisfies Faraday s Law as long as E B c Ampere Maxwell Law c is the speed of light We can then write Emax c Bmax k B Bmax sin kx t Bmax cos kx t t t Physics for Scientists Engineers 2 We can relate the angular frequency and the angular wave number k as 2 f f c k 2 Taking our assumed form for the electric and magnetic fields we can execute the derivatives March 21 2005 Which gives us 11 March 21 2005 Physics for Scientists Engineers 2 12 Ampere Maxwell Law 3 The Speed of Light We can rewrite the previous equation as Our solutions for Maxwell s Equations were correct if E E 1 c and B B 0 0 c Emax k 1 Bmax 0 0 0 0 c We can see that This relationship also holds for the electric and magnetic field at any time c E 1 B 0 0 c 1 0 0 If we put in the values of these constants that we have been using we get 1 8 1 0 0 c c Physics for Scientists Engineers 2 c Thus the speed of an electromagnetic …
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