Question

10. The expression of the magnetic field associated with an electromagnetic wave in
vacuum is given by
B = (100T) y sin (210 x 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.
(1x​

Answer 1

Answer:

As per the given question,

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

8

t+kz)

Wave number is given = K

frequency = 10^8 Hz10

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=\dfrac{E}{B}C=

B

E

E=cB==3\times 10^{8}\times100y =3\times 10^{10}N/cE=cB==3×10

8

×100y=3×10

10

N/c