3. (An EM wave with E⃗ parallel to B⃗) We have seen that the...

3. (An EM wave with \vec{E} parallel to \vec{B}) We have seen that the \vec{E} and \vec{B} fields of an elementary PW (single \vec{k}, \omega) in simple media are perpendicular to each other. However, consider the following proposal for the real parts of the fields of a PW, where \vec{E} and \vec{B} are parallel: Re \vec{E}(\vec{r}, t) = E_0 \sin \omega t (\sin kz \hat{x} + \cos kz \hat{y}) Re \vec{B}(\vec{r}, t) = \frac{E_0}{c} \cos \omega t (\sin kz \hat{x} + \cos kz \hat{y}) (a) Show that the fields satisfy Maxwell's equations in free space, provided that k = \omega/c. (b) Show that the fields may be written as the superposition of two counterpropagating, circularly polarized elementary PWs whose \vec{E} and \vec{B} fields are perpendicular to each other. (The superposition in this problem is an example of a standing wave.)

Transcript

00:01 These e and b fields as a function of time and space we want to first show that their magnitude is the constant value e not for the electric field at all time and for the b fields the magnetic field it's b not at all time so to do this we just have to take the magnitude so e of i should say z t is so square root of the square of the x plus the square of the so the i direction of the j direction so then we have e knot goes k z minus omega t squared plus e knot sine k z minus omega t squared and it may or not be apparent then that we could factor out this e knot so we get the e knot that comes out of the square root and then we just have this cos squared of k z minus omega t plus sine sine squared of k z minus omega t and um by the pythagorean theorem this equals one so then we just get that this is equal to e not and the same thing works out for the b field z t equals square root we have a negative sign here but it um you're squaring it so that goes away negative b not sign k z minus omega t squared plus b not kos k z minus omega t all squared that works out just b not as well because again since the negative goes away pull the b not out then you have the sine squared plus co squared that equals 1 since they're at the same angle inside the parentheses there next we want to show that these two fields are perpendicular we can do that by showing that e .b equal 0 that that would that's the case if two vectors are perpendicular so taking the dot product of these two vectors we get e .0.

02:37 K z minus omega t times negative b not sine k z minus omega t so that's the x components multiplied together and then plus e not sign of k z minus omega t and times b not cos k z minus omega t and there we have this negative sign that's difference here it doesn't matter the order of cosine sign both have both terms have that so this does in fact equal zero and then to compute the pointing vector this s vector here we need take the cross product so our our cross product then we just have the we have i and j components so i'm actually gonna i'll just work out what that's going to be in terms of the x and y there i should say i and j components up here so i j k i'll write the determinant matrix e i i i i i i x x x e y z and b x b y b z and we're only going to end up with components in the z direction since both of these are and the ij xy directions.

04:28 So we know this is just going to be, or yeah.

04:32 So these are zero.

04:35 So we just get the k, we get k times x, b, y, minus ey, bx...

Question

Please give Ace some feedback