Consider an electromagnetic wave with frequency 1 ghz that...

Consider an electromagnetic wave with frequency = 1 GHz that propagates in the z-direction of a right-handed coordinate system (i.e., the coordinate system considered in the lecture). E_x = E_(0x)cos(ωt - kz - φ_(x)) E_y = E_(0y)cos(ωt - kz - φ_(y)) The total electric field vector is given by: vec(E) = E_x hat(x) + E_y hat(y) Consider E_(ox) = E_(oy) and take φ_(x) = 0. Plot as a function of time: (a) E_x (b) E_y (c) the angle between vec(E) and the x-axis (d) angle between vec(E) and the y-axis for each of the following cases: (1) φ_(y) = 0° (2) φ_(y) = +180° and -180° (3) φ_(y) = +90° and -90° (4) φ_(y) = +45° and +45° Consider an electromagnetic wave with frequency = 1 GHz that propagates in the z-direction of a right-handed coordinate system (i.e., the coordinate system considered in the lecture). EEcoskz EEcoswkz The total electric field vector is given by: EETED Consider Eox = Eoy and take O. Plot as a function of time: (a) Ex (b) Ey (c) the angle between E and the x-axis (d) angle between E and the y-axis for each of the following cases: 1 = 0 2 = +180 and -180 3 = +90 and -90 4 = +45 and +45

Transcript

00:01 For this question, we're given an expression for the electric field e in free space.

00:06 We're told it's equal to e not times sine kx minus omega t in the j hat plus e not cosine kx minus omega t in the k hat.

00:14 And for the magnetic field b, it's b not cosine kx minus omega t j hat minus b not sine kx minus omega t.

00:22 So to avoid having to write kx minus omega t since the sign and cosine terms all contain that, i'm going to do a variable switch.

00:29 I'm just going to call delta kx minus omega t.

00:32 So anytime i write delta, that's what that means.

00:34 So it's just avoiding having to write that over and over again, since we're writing a lot here.

00:38 For part a, it says show that e and b are perpendicular to one another at all times.

00:43 Okay, so to do that, all we need to do is show that the dot product between the two is equal to zero.

00:48 If it is indeed equal to zero, then we know that they're perpendicular to one another at all times.

00:53 So we do e .d .b.

00:55 So when you do the dot product, you can only multiply light components.

00:58 So the i -hat and the j -hack components from e and b get multiplied together.

01:03 So what we have here is e -not times the sign of kx minus omega -t, which i said was delta, times b -not times the cosine, b -not, not just b -not, b -not, times the cosine of kx minus omega -t, delta, minus e .0 times the cosine of delta times b not times the sine of delta.

01:45 So essentially we're taking e sine delta b, b, b, not cosine delta, and subtracting it from e .0, sine, delta, b, not cosine delta.

01:55 Since everything's being multiplied, it doesn't matter the order of multiplication.

01:59 These two terms are equal to one another, so this comes out to equal zero.

02:02 And since they're equal to zero, we can say they're perpendicular to one another.

02:07 So we can say thus they are perpendicular to one another.

02:22 You can box that in as a solution for part a.

02:27 Part b says to for the e and b waves in plane that are parallel to the yz plane so that the waves move in a direction perpendicular to e and b.

02:38 So the direction of propagation is perpendicular to e and b.

02:42 So to do this we need to show that the cross product to the pointing vector is perpendicular to e and b.

02:50 It's going to need a little bit of space.

02:52 So let's start a new page here.

02:53 So the pointing vector s is defined as 1 over mu not.

02:59 Again, the pointing vector is the direction of propagation.

03:01 It's 1 over mu not, e not cross b not, or excuse me, e cross b, since we're considering the vectors in this case.

03:16 Okay, so we can use the matrix table here.

03:20 Here to more easily do this cross product.

03:24 So this is equal to, we have one over mu not, since that's just a constant, we'll pull that out front.

03:30 So the directions we have here are i -hat, j -hat, and k -hat.

03:42 Ok? well, there is, for both e and b, there's no i -hat direction.

03:48 The j -hat direction here is for e, is e -0 times the sign of delta.

04:01 For the b field, magnetic field, it is b0 times the cosine of delta.

04:11 Let me make sure i got enough room to write the k hat.

04:13 Let me get rid of this to move that over a little bit.

04:28 Okay, so the k hat for the electric field is e0 times the cosine of delta.

04:38 And for the electric field, or for the magnetic field, it's minus b.

04:41 X0 times the sign of delta okay so if we carry out this operation here we find that this is equal to 1 over mu not still times okay so the first one is minus e not times b not and since both these so what we're doing here is we're multiplying this together here those two that i just crossed in red so this is minus e not times b not times the sign squared of delta, since they both have the side of delta term.

05:33 And then we're going to subtract from that the cross multiplication in this direction.

05:39 So this is minus.

05:43 Again, you have e .0 times b not times the cosine squared of delta.

05:56 And this is in the i -hat direction.

05:58 The reason it's in the i -hat direction is because the cross -product between j and k give us i -hat.

06:03 Now, if you did the same thing for the j -hat and the k -hat, they would be multiplied by that i -hat.

06:07 The i not terms, but the i not terms are zero.

06:11 So this would be plus 0 j hat plus 0 k hat.

06:18 So we just have i hat directionality here.

06:22 So we can pull the e knot and the minus e knot and b not out front...