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.

Expert's answer

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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