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.