Question

The expression of the magnetic field associated with an electromagnetic wave in vacuum is given by (100T) ˆ sin(2 10 + ) 8 B  y   t 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.

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