00:01
Hi there, so for this problem we are told that the magnetic field component of a plane wave in a loose and a lossless dielectric environment is the following expression 30 times the sign of 2 times pi times 10 to the 8 times the time and this minus 5 times x.
00:33
So in here you can recognize the angular frequency and also the what we are going to call beta.
00:50
So for part a of this problem, we are asked to calculate, well, if mu subart is a given value of 1, we need to calculate.
01:14
Now, there is a relationship between beta, the angular frequency, and epsilon sub r.
01:24
And that is that beta is equal to the angular frequency divided by the speed of light, and that times the square root of epsilon sub r.
01:35
So now, what we need to do is to simply solve for epsilon subart.
01:47
So in here, we will have that that is beta times the speed of light divided by the angular frequency.
01:55
So to get rid of this square root, we just elevate both sides to the square.
01:59
So we will have beta to the square times the speed of light to the square.
02:03
This divided by the angular frequency to the square.
02:06
And now we just simply substitute all of the values that we are given.
02:10
So that will be phi to the square times the speed of light that we know a meters per second is three times.
02:16
Times 10 to the 8 meters per second, that to the square.
02:20
This divided by the angular frequency, and the angular frequency is 2 times pi times 10 to the 8, and that to the square.
02:32
So, using our calculator, we obtain a value of 5 .69 .93.
02:43
So that's a solution for part a of this problem.
02:48
Now, for par b, we are asked to calculate the wavelength and the wave velocity.
02:55
So for par b, we want the wavelength and the wave velocity that we are going to call as u.
03:04
Now, with that said, we know that the wavelength is just equal to the angular frequency.
03:13
Well, sorry, is equal to 2 times pi divided by beta.
03:19
So that will be 2 times pi divided by the value of beta, which is 5.
03:25
So from here we obtain a value of 1 .266 meters.
03:35
So that's a solution for the wavelength.
03:38
Now for the speed, we will have that that is the speed of light divided by the square root of musupr times epsilon to r.
03:49
We take the square root of that.
03:50
So using the values that we know, the speed of light is 3 times 10 to the 8 meters per second.
03:56
And this divided by the square root of the product between musupr, which in this case, well, we are given that value, which is 1.
04:10
Yes, 1.
04:11
1 times the value that we just obtained for epsilon sub r, which is 5 .69 .93.
04:24
So using our calculator, we'll turn that the velocity or the speed is equal to 1 .2 ,500, sorry, 1 .257 times 10 to 8 meters per second.
04:47
So that's a solution for par b.
04:50
Now for par c, we are asked about to find the wave impedance.
04:59
So the wave impedance that we're going to call new is equal to new sub 0 times the square root of mu subart divided by epsilon subart...