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 or 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 isn't energy, but it was a related units or measurement called intensity now and 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 over A and that's still true for electromagnetic waves. But now we're gonna see some new equations here that relate the intensity of elect magnetic waves to the magnitudes of the electric and magnetic fields here. So there's a couple of new equations that you're gonna see. Um And I'm just gonna go ahead and list them all for you. The first one is that is that intensity I is equal to one half C epsilon, not times E max square. 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, the other one that you'll see probably pretty commonly is one half C over mu knots times B max squared. 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 one half C. And then basically both of them have some kind of a maximum value squared whether it's E or B. And the last thing here is that E epsilon usually goes with E and U usually goes with B. So I think this is a really great way to memorize these equations. Um It also sort of looks like other energy like equations where you have one half times something times something square. All right. Now, the units, just as we went learned in, um when we talk about mechanical waves is gonna be watts per meter squared. And basically, there's two types of problems when it comes down to this intensity stuff. There's 11 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. Um 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 gonna be equal to four pi R squared. Because basically, if you draw out some little distance R, then the surface of this area basically just be become the surface area of a sphere. However, if it emits directionally like a flashlight, then you cannot assume that this area is equal to four pi R squared. And you're gonna have to go figure it out. That's basically the hardest parts of these kinds of problems is figuring out which one you're dealing with. All right. Now, I'm gonna 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. All right. So we have an incandescent light bulb that emitting at 50 watts uh 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 gonna 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. Ok. So in this first part here, I want to calculate. I, that's basically what I'm looking at here. Ok. 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 ES and the BS. Which one do I use? Well, if you take a look here, um, this unit, this 50 watts is actually the power of the uh light bulb. So that's my p so I've got that and then we know that the uh that the uh 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. All right. So if you take a look, uh let's just say that we're gonna draw this distance and we're gonna say that R equals 5 m and basically the area, the surface area here uh just becomes uh four pi R squared. All right. So we, we, we're gonna have that power is equal to 50 divided by uh area, which is gonna be four pi times five squared. If you go to work this out, you're gonna get a 0.16 watts per square meter. And that is your final answer for part A All right. Now, we're gonna 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. All right. So we're gonna calculate the electric field. So we 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 gonna see this equation here. So we're gonna have one half C epsilon, not E max squared. Since we're looking for the electric field, we're gonna use this version instead of the B max version. All right. So basically, what I'm gonna do is I'm gonna say that I is equal to uh this is still equal to P over A but now it's equal to one half C epsilon knots times E max squared. Now, I'm really looking for, what is this E max over here? So I can just rearrange my equation. All right. So I'm not gonna use this P over A because I actually already figured out what this I is equal to, right? It's just equal to this number over here. All right. So I kind of have to don't write that. So this is just a little 0.16 and this is gonna equal uh Let's see, I'm gonna move all this other stuff over to the other side. When I move this to this side. Basically, what happens is they pick up a factor of two on the outside over here and then I have to divide by C and epsilon. Knots. So this is equal to my E max squared. 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 max is equal to the square root of two times 0.16. And now I'm gonna plug in my numbers. This is gonna be three times 10 to the eighth and then epsilon knot, that's just a constant that we've got over here in this table right here. So we've got eight points. I'm just gonna use two decimals because it doesn't really matter. Uh 8.85 times 10 to the minus 12. When you go ahead and plug all of this into your square roots, what you're gonna get here uh is you should get a E max of 10.98 and remember that the units for electric field are newtons per Coolum. 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 gonna calculate now the R MS value of the magnetic field at this point. And that actually gonna 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, we kind of use that in, in thermodynamics a little bit uh instead of maximum values. And if that ever happened, there's actually a pretty straightforward relationship between R MS and max values. And it's just that uh R MS is equal to the maximum divided by the square root of two. So your er MS is gonna be em max over root two, your BR MS is gonna be B uh max divided by route two. All right. So you can always combine these equations with your other intensity equations and everything works out. All right. So just be really tricky, just really, really careful that you know which one that you're solving for. Because a lot of questions will try to throw you off and ask for er MS or B Max or something like that. So just be really careful that you're solving the right thing here. OK. So in this last part we're looking for is we're actually looking for BR MS. All right. So I know that Brns is equal to B max divided by root two. The problem is I don't know what B Max is, but it can calculate it because I know that B max number is always equal to uh is always related to um E max by the speed of light equation. So I can always calculate what B max is because I have that E max is equal to CB. So therefore your E max, which is your 10.98 divided by three times 10 to the eighth is equal to your B max. And that's gonna equal, uh that's gonna be 3.66 times 10 to the minus eight and that's in Teslas. So basically, what I can do here is I can say I'm gonna bring this all the way out here and I'm gonna say that B BR MS is equal to 3.66 times 10 to the minus eight divided by root two. And your final answer is going to be, that's gonna be uh 2.59 times 10 to the minus eight in Tesla's all right. So that's your final answer anyway, let me know if you have any questions and I'll see you in the next one.

2

example

Example 1

Video duration:

6m

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All right, folks, let's exert this practice problem here. So we're told that the average intensity of sunlight at the earth's surface is about 1400 watts per square meter. Now, remember that's I, I equals 1400. Now, in this problem, we're asked for two things, there's two parts. The first part asks us to calculate the amplitudes of the electric and magnetic fields for this wave. Now, remember whenever they're asking you for the amplitude, you should always think that's the maximum or the magnitude of the fields. So amplitude always means either B Max or E Max. And in this case, they're actually asking us for both of them. So let's just go ahead and get started with that, right. So, in part A they're asking us for Em Max and B Max. Now, whenever you're relating, uh now whenever you're asked for em, Max and B Max and you have the intensity, you can always use this equation over here. This I is equal to or power times area, but it's also related to em Max and B Max. And there's two separate equations for it. So we're just gonna go ahead and get started there. Uh the first one we're gonna tackle is the E max one. So remember that the I is equal to power divided by area. But in this particular part, in part, a, we don't care about power or area. So this is not the expression we're looking for, but we do want to relate this to the E max. And remember this equation here says that I is equal to one half times C epsilon, not E max squared. All right. So we have what the intensity is. And if we, if we want to calculate E max, then all we have to do is just rearrange this equation because remember C and epsilon are just constants. So that's what we're gonna do over here. We're gonna take this, we're gonna pull it over to the other side. And we're gonna say that I divided by one half C epsilon knots is equal to E max squared. Now, by the way, you could move this one half over to the top if you want. But you could also leave it in the denominator which I did. Now, the last thing we need to do here is this is gonna calculate what emac squared is. But we want em ma we just have to take the square roots. So this is just be the square roots. And now we can just start plugging in some numbers here. So we have the square roots of 1400 divided by one half times three times 10 to the eighth and then epsilon knot, remember it's just a constant and all the constants are over here on the right side of this table is 8.85 times 10 to the minus 12. All right, now, you don't have to put the units because the units will just cancel out and you'll get uh something that is in newtons per Coolum. You can kind of just trust me on that and this is gonna equal E max. And so this is gonna be if you actually plug this in 1027 newtons per cool loan. All right. So this is the final answer for your electric field. Now, that's one half of part A because we also want to calculate what is the magnetic field B max? Now, B max actually, now that we figured out E MA you can actually solve it one of two ways here. All right. So I put point out here that Eb Max can be solved using two ways. Now, one of the ways you could do it is you could basically do exactly what we did for the electric field, but set the I equal to one half C over Muno B max square. All right. So basically what we could do here is we could say that B max is equal to one half C over mu knots times. Um We can actually sorry, this is gonna be I, so I is equal to one half C over not B, not B max squared. And you could basically go ahead and solve for this B max using this method or now that we've actually solved for the E max. Remember you can always relate to em Max and B max by using this equation over here that E max equals C times B max. All right. So, or we can put that E max is equal to C times B max. Now that we figured out what emacs is, you can actually just plug it into the equation. All right. So basically, whenever you figure out one of them, you can always figure out what the other one is by sort of taking this shortcut equation over here. All right. So that's what we're gonna do because it's gonna be a little bit faster. So this is gonna be uh emacs divided by C which will equal B max. And if you go ahead and plug this in which you're gonna get is 1000 and 27 divided by three times 10 to the eighth. And when you work this out, what you should get uh is you should get 3.42 times 10 to the negative six in Tesla's. Now, what I want you to do is I want to pause the video if you're just, if you don't trust me and you can actually work this out by using this equation over here and you should get that B max is equal to the same thing. All right. So let's go ahead and move on now to Part B. All right. So that was part A now part B asks us to find what now part B says, if the earth sun distance is approximately 1.5 times 10 of the 11th meters, then what's the power output of the sun? So now we're looking for the average power output? All right. So basically what that means here uh is that we're looking for P and by the way, every time they say power, the average power output output, that P average really just means P. All right. So there's no really, there's no difference between average power and power. It's all just the same P, all right. So then how do we calculate this P over here? Now remember that power is always related to intensity via this, these two equations over here I is equal to P over A. So we have what the intensity is. So we basically say that I is equal to P over A. Now, the one thing we need to do is we just need to figure out what kind of sort of source are we dealing with? Here? Are we dealing with the source that radiates outwards radially or is it directional? Remember if we're talking about the sun here, the sun is a spherical object and it radiates light everywhere. It's not like a laser beam or a flashlight. So basically what this means here is that your A is a sphere which equals uh four pi R squared. All right. So basically what we can do here is we can say that the power is gonna be equal to intensity times A. And this A here is just gonna be four pi R squared where this R here is the R from the earth to sun distance, right? So this is to be uh earth sun distance R squared. And basically, we can go ahead and plug this in. So this is gonna be the intensity, which is 1400 that's the intensity in watts per square meters. We don't have to do any conversions and then we can do four pi times the radius between earth and the sun, which was given to us, which was 1.5 times 10 to the 11th and it's in meters. Uh So we don't have to convert anything like that. And when you go ahead and plug this in what you should get for the total power output of the sun is 3.96 times 10 to the 26 in watts. Now, you could actually just go ahead and search this up online and you'll find that this actually is the intensity of the sun or, or this is sometimes also called the luminosity of the sun. All right. So that's it for this one. Folks, let me know if you have any questions?

3

example

Example 2

Video duration:

7m

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All right folks. So let's take a look at this example problem here. So we have a radio tower that's emitting some kind of a radio signal with an average power output of 50,000 watts. So in other words, that's our P or P average, this equals 50,000. All right. Now, this radio tower emits equally in a hemisphere above earth's surface. Basically what happens is if you go to kind of imagine that the ground is over here, the signals can't penetrate through the ground. So it's sort of equally. Uh so it radiates in a hemisphere, not a perfect sphere. All right. Now, what we want to do is we want to calculate the intensity that detected by this satellite that's passing overhead uh at a height of 100 kilometers. So basically what this means here is that from the radio tower up to the satellite here, this is gonna be my H which equals 100 kilometers. All right. So let's start with that first part um which is the intensity. So remember intensity, if we want to calculate the intensity, we just have this one big long expression to do this. This is just power divided by area. And then we also have some expressions that deal with the electric and magnetic fields. Now, in this problem, we have no information about E MA er MS B max or BR MS or anything like that. So we're probably not gonna use that um those expressions here. Really, this whole problem can be solved just by using I equals P over A right. We have the power, uh we have the power over here and presumably we know something about the hemisphere. So we can calculate the area. So that's what we need to calculate the intensity. All right. So if you take a look, this, this power is just gonna be 50,000 watts over here divided by what's the area. So what's the area of this um of this source or what's the area that the signal is sort of spread out through rounds? Remember if you can assume that it radiates radially spherical, then the area is equal to pi R or four pi R squared, but this is not a sphere. So what happens here is that the area of a hemisphere over here is really just equal to one half of the area of the actual sphere. So in other words, it's gonna be one half of four pi R squared. So basically what this means here is that the area of a hemisphere and what we're going to plug into our equation is actually two pi R squared All right. So that's the tricky part here. Just kind of figuring out what the area is. You can't just assume it's four pi R squared, but that's where we're gonna plug into our area formula. All right. So this is gonna be two pi, now, what's the radius? So the radius really is just the distance between the radio tower and the satellite. That's H but that's also equal to R. So just be really careful here because you don't wanna plug in 100 you have to convert it to um si units. This is actually just gonna be 100 you can just tack on 30 or 100,000. Or you could also just write this as one times 10 to the fifth. Uh And then you just wanna square that. So one time instead of the fifth squared and what you'll get for the intensity, the intensity is equal to uh this is gonna be 7.96 times 10 to the minus seven. And remember the units for this are in watts per meter squared. That's always the units for intensity. All right. So that is the answer to the first part here. That's the intensity. Now let's move on to the second part. The second part now asks us, let's see if the power or so it asks us to calculate the power that's received by the satellite's circular radio antenna with a radius of 0.5 m all Right. So here's what's going on. I sort of want to draw this out for part B the satellite here has a little, little dish, um, like this, uh, that's looking for signals or whatever. And I'm gonna sort of draw this out really, really quick here. Uh You can sort of imagine that this sort of dish has a little cone thing like this. Uh And it's taking in signals, basically what happens is it's taking in the intensity from this radio tower, all those signals here and there's gonna be a power now that gets, um now that actually gets focused on this little area here. OK. So here, what we have is we actually want to figure out what is the power of the antenna or what's the power that the antenna receives. Now, one of the things I want to point out here is the power is not just gonna be the power that the radio antenna broadcasts at its 50,000 watts because remember as these signals get out towards this and spread out around this sphere, they lose intensity, right? So that power basically dissipates over a larger area. And so it's definitely not gonna be 50,000. So how do we go about solving this? Well, really if you think about this, now you're kind of just doing the reverse of what you did in part A. So what happens here is we're still gonna use I equals P over A. But now we actually have what the intensity is at this distance here. Now, we want to figure out what is the power that the antenna receives? All right. So to calculate the power that the antenna receives, we're also gonna need to know the area of the circular antenna. All right. So you're kind of just taking this equation and you're flipping it around. First, we used the power of the radio. So this is the power radio over area of the radio tower, that was the intensity and now we're actually sort of doing the opposite. Now, we know what the intensity is. We wanna calculate what's the power received by this antenna? All right, just wanna make that super clear because I it can be kind of confusing. All right. So um let's go ahead and calculate what the power is. Um basically just move this over to the other side. So power of the antenna is gonna be I times the area of the antenna. All right. So the area of this antenna, now this area of this antenna is not just gonna also, it's also not gonna be the same area that we calculated over here. That area was the area of the whole entire hemisphere. So what's the area of this antenna? We're told that the radius of the antenna. So I'm gonna say the R antenna is equal to 0.5. So in other words, the area of this antenna is actually just gonna be pi times R antenna squared, right? Because it's basically just the area of the circle like this. All right. So this, by the way, um uh if, if you um actually, yeah, so, so that's the R antenna square. All right. So that means that this is just gonna be the power of the antenna is equal to the intensity, the intensity is 7.96 times 10 to the minus seven. And now we're gonna multiply this by the area of the antenna which by the way is just pi times 0.5 squared. All right. So this is just, this just means that the power that the antenna receives is equal to 720. I'm sorry. Uh this is gonna be uh 6.25 times 10 to the negative seven and this is watts. All right. So that is your final answer. OK. So just to sort of recap we did in part a we use the power of the radio that the actual radio tower produced. And then we use the, the area of the hemisphere that it broadcasts over to figure out what the intensity is at this specific point. And this intensity was 7.96 times 10 to the minus seven. Then what we did is we sort of reversed it and we said, well at this intensity here, what's the power that now gets received by a panel or a radio dish? That is approximately 0.5 m wide. So we actually had to reverse the equation. Now we have to solve for a different p than we used up here. All right. So hopefully that made sense. Just wanna make that super clear. Uh Let me know if you have any questions in the comments and I'll see you in the next video.

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