Question #115210
The expression of the magnetic field associated with an electromagnetic wave in
vacuum is given by
B = (100T) y sin (2πx 10⁸t + kz)
Determine the wave number, frequency and the direction of propagation of the wave
and the magnitude and direction of the electric field associated with it.
1
Expert's answer
2020-05-12T09:57:41-0400

As per the given question,

B=(100T)ysin(2π×108t+kz)B=(100T)y\sin(2π×10^8t+kz)

Wave number is given = K


frequency =108Hz10^8Hz

Direction of the propagation is in along the z axis

We know that the relation between the electric field (E)and the magnetic field(B) and the speed of light(c) is

c=EBc=\frac{E}{B}


E=cB=3×108×100y=3×1010N/cE=cB=3×10^8×100y=3×10^{10}N/c


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