Hello, everyone, and welcome back. So when we studied mechanical waves, we saw that some questions would not only ask you to draw the functions but also ask you to write out their wave functions. We're going to see the exact same thing for electromagnetic waves. And so what I'm going to show you how to do in this video is how to write the wave function for electromagnetic waves. We're going to see a ton of similarities between how we did this for mechanical waves, but we'll also see some differences that you'll need to know to solve problems. So let's go ahead and just jump right in. So remember that electromagnetic waves are transverse waves which means that there's something going up and down. Now, for mechanical waves, it was the string itself. You whip the string up and down, and the string is going up and down. For electromagnetic waves, it's a little bit different because it's actually the electromagnetic field itself that's sort of oscillating. It's going up and down like this. Now the magnetic field was sort of going forwards and backward like that and we saw that these things were perpendicular, but they're still transverse. There's still something sort of going up and down or side to side or something like that. Alright? So this was the e field, and then this was the b field. Alright? So because these are also transverse waves, then we can also use sine or cosines to describe them. So we can use sinusoidal functions. Alright? And we're going to have to write their wave functions. Alright? The only difference between electromagnetic and mechanical waves is that because electromagnetic waves are made up of 2 oscillating fields, we have an e field and a b field, then we're going to have to write 2 wave functions to describe them. Alright? So whereas for mechanical waves we only needed one equation, for electromagnetic waves, we'll need one for the electric field and then one for the magnetic field. Otherwise, what we're going to see is that the structure of these equations is actually very similar. Alright? So let's talk about that. The structure of the mechanical wave equation was we had an amplitude and we had a sine term and then we had a kx−ωt. What we're going to see for electromagnetic waves is something very similar. We have a sine and a kx−ωt. And by the way, some of your books might actually use cosine instead of sine. If your book used a cosine for this function, then they're going to use a cosine for the electromagnetic waves. But remember, you can use either one. The difference really just has to do with where the wave starts. Like, if the wave started over here, we would use a cosine, but that's all arbitrary. Right? So it's totally fine if you see a cosine. Alright? Now this front term here for the mechanical wave was the amplitude. And the, sort of analogy for mechanical waves, it was basically how high the string actually was going when you whipped it up and down. For the electromagnetic waves, it's a little bit different because there's nothing really going up and down. What's going on here is that the electric field strength is changing. So what happens here is that this maximum value, we're not going to use amplitude, we're actually just going to call this E max. And similarly, over here, this is going to be B max. This is sort of like the amplitude analog for an electromagnetic wave. Alright? So what goes on here in this front term isn't a, it's just E max and B max. Alright. That's really the only difference to these equations. Everything else structurally is the exact same. For example, this k term over here we found was the wave number which was1λ. It's the exact same thing. It's just that you have a different λ because light waves are different than mechanical waves. And then for this omega term which was the angular frequency, it's going to be the exact same sort of variable here. It's going to be angular frequency, just your f is going to be a little bit different. Alright? So the last thing I want to point out here is that the electromagnetic wave, the E and the B field, are always what we call in phase. And this just means that they have the same kx−ωt term, which is actually really useful for us. Because basically what it means is that once you figure out what the kx−ωt is for one of the equations, then you figure it out for both of them because it's going to be the same for both of them. Alright. So it's going to be the same kx−ωt. Once you've solved it once, you've solved it for both of them. And by the way, this, phrase in phase just means that they reach their minimum zero and maximum values simultaneously, which we've also seen from the graphs. They hit their 0 points at the same point. They hit their maximums at the same point respectively in their cycle and then they just repeat the whole process over and over again. Alright? So let's just go ahead and jump into a problem right here. We've got an infrared laser that emits a 10 micrometer wavelength. This is going to be our λ, in the x direction. So we've got that the e field is parallel to the y axis and it's got a max value of 1.5. By the way, when it says that the e field is parallel to the y axis, what you'll see here is that this this e field is written with a y. That just means that it's sort of parallel oscillating in the y axis. Alright. So that means we're going to have the same thing over here. The e is oscillating in the y axis. So Eyx,t and then on the b fields, we're going to have parallel to the z axis. And what we actually want to do here is we actually want to write out what their wave functions are. Alright? So that's what we're tasked to do here, write the wave functions of e and b. Now the first thing you might be wondering is what do I actually pick? Do I have to pick a sine or do I have to pick a cosine? Which do I use? And so it actually brings me to the last point I want to make in this video, which is that if problems don't indicate a wave's start point with either the graph or the text, then you could actually use either sine or cosine. Sometimes you get a problem in which it doesn't really specify which one you're supposed to pick, and it's because it's kind of arbitrary. Right? So that's not as important as it was for mechanical waves. You can pick sine or cosine for this particular problem. So I'm just going to go ahead and pick sine. So this is going to be E max, and then I'm going to have a sine of kx−ωt, and then I've got B max. This is going to be sine of kx−ωt. Alright? And by the way, we know this is going to be a minus sign like it was for mechanical waves because it's traveling in the x direction and we can assume that that's just positive. Alright. So if we take a look at our variables here, we actually have what the maximum value of the electric field is. So that's what we've got here. What we really need to do is we need to figure out what the k and the omega are because that's what I need to actually write out the full equation. Alright. And you'll notice here that I also don't have what the B max is, but I can also figure that out. Alright. So the first thing I'm going to do is I'm actually going to look at this k term over here and see if I can figure that out. So how do we figure out k? Well, if you remember, k stands for the wave number and there's a special equation for that which is 2π/λ. So this is going to be k=2π/λ, and I actually have what λ is. Well, λ is just 10 micrometers. So this is going to be 2π∗10·10−6. Alright? So if you go ahead and work this out, what you're going to get, is you're going to get, let's see. This is going to be this is going to be 6.28 times 10 to the minus 5, and, we don't actually need the units for that. Alright. So we've got what the k term is. Now let's go ahead and look at the second term here. We need the omega term. I'm going to go ahead and work that out over here. So we need omega. Now remember, omega is the angular frequency, which is equal to 2π∗f, but it's also equal to something else, which is actually much easier to get, which is just v∗k. So in other words, remember that v for electromagnetic waves is actually going to be c, so this is also going to be equal to c∗k. Now just because we figured out what k already is and we know what c is, we can actually just figure it out more directly this way. So this is just going to be 3∗108 and this is going to be times k, which is 6.28∗10−5. Alright. So if you go ahead and work this out, what you're going to get here for omega is you're going to get, this is going to be let's see. What I get is 1 point oh, I'm sorry. This is, times oh, this shouldn't be negative. This actually should be a positive. Oops. So that should be a negative sign there. It should be positive. And what you're going to get here is you're going to get, 1.88∗1014, and that is going to be the angular frequency. Alright. So that's your angular frequency. The last thing I need to do is now I just need to figure out what B max is. That was my last unknown variable. So how do I figure that out? Well, I can always just relate B max or any of the magnitude of B to E by using the equation E=cB over here. So I have what B max is equal to sorry, I have what E max is equal to. So in other words, Emax/c=Bmax. So that means that 1.5∗106, this is going to be 5 divided by 3∗108 is going to give me my my B max which is going to equal 0.005, otherwise known as 5∗10−3 Teslas. Alright. So what we can do here is now that we have all of our variables, we can write out the final expressions. So that Ey (x,t ) ⇒ 1.5∗106 times the sine. Now we've got the k term which is 6.28∗105, minus and that's going to be the x. Don't forget the x, minus 1.88∗1014 t. And then we've got that Bz of x and t is going to equal, 5∗10−3. This is going to be sine, and this is going to be 6.28∗105 x minus 1.88∗1014 t. Remember we can just plug in the same exact values for this kx minus omega t term. Alright? And this is your final answer by the way. This is your

# Wavefunctions of EM Waves - Online Tutor, Practice Problems & Exam Prep

### Wavefunctions of EM Waves

#### Video transcript

Electromagnetic waves produced by X-ray machines typically have a frequency of approximately $3.5\times10^{16}Hz$ . What is the wave number of these waves?

$2.20\times10^{17}m^{-1}$

$1.80\times10^{-16}m^{-1}$

$1.17\times10^8m^{-1}$

$7.33\times10^8m^{-1}$

### Example 1

#### Video transcript

Hey, everyone. Let's get started with this example problem here. So we're talking about what the wave function of the electric field of this electromagnetic wave is. It's this \( E(x,t) = 54 \sin(2.2 \times 10^7 x - \omega t) \). So what we can see here is, from our formula, remember this is the sort of general equation for an \( E_{\text{max}} \) and this \( 2 \times 10^7 \) is actually our \( k \), but we don't know what \( \omega \) is. Alright. So we have \( E_{\text{max}} \) and we also have \( k \). So the first thing we need to do is to find the amplitude of the magnetic field. Remember, whenever they're asking you for the amplitude, they're just asking you for the maximum value here. So they're asking you for what is \( B_{\text{max}} \). Alright? So how do we calculate \( B_{\text{max}} \) here? Well, whenever you want to figure out either the strength of one of the fields, whether it's the electric field or the magnetic field, you can always relate it to the other one by using \( E = cB \). Now \( E = cB \) works for anything, but it could also be \( E_{\text{max}} = c \times B_{\text{max}} \). Right? So that works always. So in other words, the way we figure this out is by using \( E_{\text{max}} = c \times B_{\text{max}} \). So we divide over the \( c \), and this really just becomes \( \frac{E_{\text{max}}}{c} = B_{\text{max}} \). Now we're just going to go ahead and plug in some variables. Now remember this is just 54, so that's what we plug in for \( E_{\text{max}} \). \( \frac{54}{3 \times 10^8} \), and what you should get is \( 1.8 \times 10^{-7} \), and that's in Teslas. Alright? So that's your final answer for part a. Let's move on to part b now.

Now, we want to figure out what's the frequency of this electromagnetic wave. Remember, frequency is the variable \( f \). So how do we do that? Well, you'll notice here the easiest, I guess, most direct way to figure out the frequency is by trying to relate it to the \( \omega \) if you had \( \omega \). Unfortunately, in this equation, we don't have \( \omega \). Remember, this is just a variable over here. It's something, some number that we just don't know. So instead of using this \( \omega = 2\pi f \), we can't use this because we don't know what omega is, but we do know what \( k \) is. Right? Remember \( k \) is just this variable over here or this number \( 2 \times 10^7 \). So instead, we can use this part of this equation here. Remember these three things are equal to each other. So, whichever two of the variables that you have, you can always use those two to figure this out. So we want to figure out the frequency and we know what the \( k \) is, so now we just use this part over here, \( 2\pi f = c \times k \). Now we just divide the \( 2\pi \) over to the other side. So this becomes \( f = \frac{ck}{2\pi} \). So I'm going to start plugging in some numbers. So this is \( 3 \times 10^8 \) speed of light, our \( k \) variable is \( 2 \times 10^7 \). Remember, this doesn't have any units; it's just a number, and then we just divide by \( 2\pi \). Now when you work this out, what you should get is you should get \( 9.55 \times 10^{14} \) and that is in Hertz. Alright. So that is your final answer for the frequency of this electromagnetic wave. So you can always basically get these variables \( k \) and \( \omega \) from these two variables over here. Alright? So let me know if you have any questions, and I'll see you in the next video.

The magnetic field of an electromagnetic wave traveling along the z-direction is described by the wave function $B\left(z,t\right)=1.0\times10^{-3}\sin\left(kz-1.27\times10^{12}t\right)$, where k is the wave number. Write the *complete* wave function for the electric field of this wave.

$E\left(z,t\right)=1.0\times10^3\sin\left(3.81\times10^{20}x-1.27\times10^{12}t\right)$

$E\left(z,t\right)=3.0\times10^{11}\sin\left(3.81\times10^{20}x-1.27\times10^{12}t\right)$

$E\left(z,t\right)=3.0\times10^5\sin\left(4.23\times10^3x-1.27\times10^{12}t\right)$

$E\left(z,t\right)=1.0\times10^3\sin\left(4.23\times10^3x-1.27\times10^{12}t\right)$