Question #104222
The expression of the magnetic field associated with an electromagnetic wave in
vacuum is given by
B = (100T) y sin ( 2π × 10^8t + 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
2020-03-09T10:58:36-0400

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 = 108Hz10^8 Hz

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=\dfrac{E}{B}

E=cB==3×108×100y=3×1010N/cE=cB==3\times 10^{8}\times100y =3\times 10^{10}N/c



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