Question
Question: The electric field associated with an electromagnetic wave in vacuum is given by \[|\vec E| = 40\cos...
The electric field associated with an electromagnetic wave in vacuum is given by ∣E∣=40cos(kz−6×108t) , where E , z and t are in volt per meter, meter and second respectively. The value of wave vector k is:
A. 2m−1
B. 0.5m−1
C. 3m−1
D. 6m−1
Solution
We can solve such questions in which equations are given and we have to find a particular given quantity. For such questions make comparison of the given equation with the standard equation of wave form. If the given equation is not in the standard form then you have to manipulate it and then compare the quantities of the equation and we are done.
Complete step-by-step solution :So the given equation of electromagnetic wave in vacuum is:
∣E∣=40cos(kz−6×108t)
And we are asked to find the value of k
The standard equation of electromagnetic wave is:
∣E\lower0.5em←∣=Eocos(kz−ωt)
Where:
Eo is magnitude of electric field
z is distance
ω is angular frequency
t is time
Now on comparing the given equation with the standard wave equation we get:
ω=6×108
Eo=40
We also know that speed of electromagnetic wave can be given by:
v=kω
Where v is the speed of light
We can also say
k=vω
Now putting the value of ω in above equation and we know that speed of light is 3×108
We get:
k=3×1086×108
k=2m−1
Hence, Option A is correct.
Note:-
Electromagnetic (EM) waves are basically changing electric and magnetic fields, transporting energy and momentum through space. Maxwell described EM waves in brief, which are the fundamental equations of electrodynamics. Electromagnetic waves require no medium to propagate, they can travel even through empty space. Sinusoidal plane waves are one type of electromagnetic waves. Not all EM waves are sinusoidal plane waves, but all electromagnetic waves can be viewed as a linear superposition of sinusoidal plane waves traveling in arbitrary directions.