Poynting Vector - General Physics - Solved Past Paper, Exams of Physics

Download Poynting Vector - General Physics - Solved Past Paper and more Exams Physics in PDF only on Docsity! 4. (25 pts) The intensity of light is the time average of the Poynting vector, ~S = (1/µ0) ~E × ~B. a) (10 pts) When writing the magnitude of the Poynting vector, what angle would you use in the cross- product, and why? The only thing this angle depends on is the direction of the two vectors involved in the cross-product. The angle you use in a cross-product is the angle between the two vectors involved. Here, that’s the angle between ~E and ~B. Both Faraday’s Law and Ampere’s Law indicate that the electric and magnetic fields are always perpendicular. Thus you’d use an angle of 90◦. This has nothing to do with the fact that this angle happens to give the largest magnitude of the cross product. b) (15 pts) Given the electric field as a wave function E = E0 cos(kx − ωt), the magnetic field as the wave function B = B0 cos(kx−ωt), and the relation between the electric and magnetic field amplitudes, E0 = cB0, show that the intensity can be written as I =< ~S >= E2 0 2µ0c All the pieces are given in the problem, just plug them in and actually do the math! We’re given the pointing vector in part (a) as ~S = 1 µ0 ~E × ~B The magnitude is |~S| = 1 µ0 | ~E| | ~B| sin 90 = E B µ0 Then we’re given wave functions for both E and B in part (b), so make those substitutions: |~S| = E0 cos(kx − ωt)B0 cos(kx − ωt) µ0 = E2 0 cos2(kx − ωt) µ0c Finally, we’re told both at the very beginning of the problem and in the mathematical notation in part (b) that the intensity is the average of the Poynting vector. So take the average of the expression above: I =< ~S >=< 1 µ0 E2 0 cos2(kx − ωt) >= E2 0 µ0c < cos2(kx − ωt) >= E2 0 2µ0c