00:01
All right, so in this problem we have a electromagnetic wave whose wavelength is 432 nanometers.
00:10
And we also know the magnitude of the magnetic field b -max equal 1 .25 microtestma.
00:21
And we also know that this wave is propagating along the positive x direction.
00:28
So in part a, we want to find out the frequency of this wave.
00:33
So according to the definition, the frequency equals c over lambda, right? so c is the speed of light.
00:41
So this equal 6 .94 times 10 to the 14th hertz.
00:48
And in part b, we want to find out the magnitude of the electric field, e max.
00:54
So e max equals c times b max, all right? so b max is the magnitude of the magnetic field, and c is the speed of light.
01:02
And you see that this value is 375 volt meters.
01:10
And part c, so we want to express the magnetic field and the electrofield in this way.
01:19
So because we already know for the electrofield, the magnitude is 375.
01:26
So it's 375.
01:28
And then we need to consider the vibration term, which is cosine kx minus omega -t, right? so k here is the wave number which can be calculated as 2 pi over lambda...