Hey, folks. Welcome back. One of the things we talked about with mechanical waves is that they carry energy, whether it was waves on a string, an ocean wave, or something like that. But that's not a property that's unique to mechanical waves. Remember that all waves carry energy. We've got sound waves. We've got light that's coming from the sun. All of those things carry energy. But a more useful measure isn't energy, but it was a related unit or measurement called intensity. Now I'm going to spend some time, but remember that intensity was basically defined as the energy per time divided by area, or you might remember that also as the power per area. Remember, energy over time is just power. We saw this equation here that intensity was just equal to p/A, and that's still true for electromagnetic waves. But now we're going to see some new equations here that relate the intensity of electromagnetic waves to the magnitudes of the electric and magnetic fields here. So there's a couple of new equations that you're going to see, and I'm just going to go ahead and list them all for you. The first one is that the intensity \( I \) is equal to \( \frac{1}{2} c \epsilon_0 \times E_\text{max}^2 \). Now you might see this written in a bunch of different ways in your textbooks. They like to throw out a bunch of different variations of this equation, but the other one that you'll see probably pretty commonly is \( \frac{1}{2} \frac{c}{\mu_0} \times B_\text{max}^2 \). If you haven't seen this, I highly recommend that you learn it this way because I think it's really easy. One of the things that's really easy to see about these equations is that both of them have a factor of \( \frac{1}{2} c \), and then basically both of them have some kind of maximum value squared, whether it's \( E \) or \( B \). And the last thing here is that \( \epsilon \) usually goes with \( E \) and \( \mu \) usually goes with \( B \). So I think this is a really great way to memorize these equations. It also sort of looks like other energy-like equations where you have \( \frac{1}{2} \) times something times something squared.

The units, just as we learned when we talked about mechanical waves, are going to be watts per meter squared. And basically, there are 2 types of problems when it comes down to this intensity stuff. There's one problem in which the source of light will emit radially in all directions like a flashlight or like a bulb or like a star or something like that. Or you might have something where it emits directionally, like a flashlight or a laser or something like that. And basically, the gist of these two different types of problems is that if you can assume the source emits equally in all directions, basically if you can assume that it's this case, then that means that the area, this \( A \) term that pops up in your area equation in the intensity, is going to be equal to \( 4\pi r^2 \) because, basically, if you draw some little distance \( r \), then the surface area of this field basically just becomes the surface area of a sphere. However, if it emits directionally like a flashlight, then you cannot assume that this area is equal to \( 4\pi r^2 \), and you're going to have to figure it out. That's basically the hardest parts of these kinds of problems is figuring out which one you're dealing with.

Now I'm going to go ahead and make a point here about the second thing, but, actually, before I do that, let's just jump right into a problem. Alright? So we have an incandescent light bulb that is emitting at 50 watts, in all directions. So because it's all directions here, that means that we can actually go ahead and use this relationship here. So that means we know what our area is going to be. It's got no inefficiencies or energy losses, and we want to calculate what is the intensity of light at some distance away from the light bulb. Okay? So in this first part here, I want to calculate \( I \). That's basically what I'm looking at here. Okay? So remember that \( I \) has a couple of different formulas. It's got power over area, and it's got all these other things with the \( E \)'s and the \( B \)'s. Which one do I use? Well, if you take a look here, this unit, this 50 watts is actually the power of the light bulb. So that's in my \( P \), so I've got that. And then we know that the light bulb emits in all directions, so we actually already know what the area is as well. So we actually have both of these, and we can go ahead and calculate this. Alright? So if you take a look, let's just say that we're going to draw this distance and we're going to say that \( r = 5 \) meters. And, basically, the area, the surface area here, just becomes, \( 4\pi r^2 \). Alright? So we're going to have that power is equal to 50 divided by the area, which is going to be \( 4\pi \times 5^2 \). You've got to work this out. What you're going to get is \( 0.16 \) watts per square meter, and that is your final answer for part a.

Now we're going to move on to the second part here, which is we're going to calculate the maximum value of the electric field at this distance here. Alright? So we're going to calculate the electric field. So we'll go back to our equations for intensity. Remember that intensity is related to the magnitudes or the strengths of the electric and magnetic fields, and we're going to see this equation here. So we're going to have \( \frac{1}{2} c \epsilon_0 E_\text{max}^2 \). Since we're looking for the electric field, we're going to use this version instead of the \( B_\text{max} \) version. Alright? So, basically, what I'm going to do is I'm going to say that \( I \) is equal to, this is still equal to \( P/A \), but now it's equal to \( \frac{1}{2} c \epsilon_0 \times E_\text{max}^2 \). Now I'm really looking for what is this \( E_\text{max} \) over here, so I can just rearrange my equation. Alright? So I'm not going to use this \( P/A \) because I actually already figured out what this \( I \) is equal to. Right? It's just equal to this number over here. Alright? So this is just equal to 0.16, and this is going to equal, let's see. I'm going to move all this other stuff over to the other side. When I move this to this side, basically, what happens is I pick up a factor of 2 on the outside over here, and then I have to divide by \( c \) and \( \epsilon_0 \). So this is equal to my \( E_\text{max}^2 \). Now the last thing I have to do here is I just have to take the square root and then plug in all of my numbers. So in other words, \( E_\text{max} \) is equal to the square root of \( 2 \times 0.16 \). And now I'm going to plug in my numbers. This is going to be \( 3 \times 10^8 \). And then \( \epsilon_0 \), that's just a constant that we've got over here in this table right here. So we've got 8.85 times \( 10^{-12} \). When you go ahead and plug all of this into your square root, what you're going to get here, is you should get an \( E_\text{max} \) of 10.98. Remember that the units for electric field are newtons per coulomb. So that is the answer to part b. Let's move on now to our final piece here, our final step, which is that we're going to calculate now the RMS value of the magnetic field at this point. And that actually brings me to the second point that I had in this video, which is that in some problems, you might be given or asked for some kind of an average value or some kind of a root mean square. Remember we used that in thermodynamics a little bit, instead of maximum values. And if that ever happens, there's actually a pretty straightforward relationship between RMS and max values, And it's just that, RMS is equal to the maximum divided by the square root of 2. So your \( E_\text{RMS} \) is going to be \( E_\text{max} / \sqrt{2} \). Your \( B_\text{RMS} \) is going to be \( B_\text{max} / \sqrt{2} \). Alright? So you can always combine these equations with your other intensity equations and everything works out. Alright? So just be really careful that you know which one you're solving for, because a lot of questions will try to throw you off and ask for \( E_\text{RMS} \) or \( B_\text{max} \) or something like that. So just be really careful that you're solving the right thing here. Okay? So in this last part we're looking for is we're actually looking for \( B_\text{RMS} \). Alright? So I know that \( B_\text{RMS} \) is equal to \( B_\text{max} / \sqrt{2} \). The problem is I don't know what \( B_\text{max} \) is, but I can calculate it because I know that \( B_\text{max} \), remember, is always equal to, is always related to, \( E_\text{max} \) by the speed of light equation. So I can always calculate what \( B_\text{max} \) is because I have that \( E_\text{max} \) is equal to \( c B \). So therefore, your \( E_\text{max} \), which is your 10.98 divided by \( 3 \times 10^8 \) is equal to your \( B_\text{max} \), and that's going to equal, that's going to be \( 3.66 \times 10^{-8} \), and that's in Teslas. So, basically, what I can do here is I can say, I'm going to bring this all the way out here, and I'm going to say that \( B_\text{RMS} \) is equal to \( 3.66 \times 10^{-8} / \sqrt{2} \). And your final answer is going to be, that's going to be, \( 2.59 \times 10^{-8} \) in Teslas. Alright? So that's your final answer. Anyway, let me know if you have any questions, and I'll see you in the next one.