In another universe, the permittivity of free space, 0 , is measured to be 3.767

Question

In another universe, the permittivity of free space,  0, is measured to be 3.767×10-12 F/m, and the permeability of free space, µ0, is measured to be 6.283×10-7 H/m. In this universe, Maxwell’s equations hold in the exact same form as in our universe, which implies that all the things covered in Ch. 31 apply, just with a different speed of light. An EM wave of frequency 5.45×1014 Hz travels in vacuum through this universe in the +x-direction. What are the speed of this wave and its wavelength? The maximum magnitude of the electric field is measured to be 2150 V/m, and it is oscillating in the z-direction. What is the maximum magnitude of the magnetic field and how is it directed when the electric field is at its maximum in the negative z-direction (E = -E0k) ? What is the (time-averaged) intensity of this wave?

(I know this has more than one question...if you think it's too much work for one question let me know when you answer it and I'll ask it again, and answer whatever and I'll give you another 350)

Explanation / Answer

1. What is the speed of this wave and its wavelength?

c=1/sqrt() = 6.500 * 10^8m/s

wavelength : c=f =c/f = 6.5 * 10^8 / 5.45 / 10^14=1193nm

2. The amplitude of the magnetic field B has the following relation with the amplitude of electric field E

E/B=c

So the maximum value of the magnetic field is

B=E/c = 2150 / 6.5 / 10^8= 3.308 * 10^-6 T

The wave is travelling in +x direction. One thing to remember is that the direction of the propagation is equal to the direction of the vector E x B(cross product) Therefore the direction of the magnetic field must be positive y-direction.

3. Time-averaged intensity : intensity at each instantaneous moment is

EB/. After the time average, it becomes EB/2 (now E and B are just amplitudes)

I = 2150 * 3.308 * 10^-6 / 2 /  (6.283×10^-7)= 5660W

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