Answer to Question #115210 in Electricity and Magnetism for Nitesh Sundas

Question

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.

Answer 1

As per the given question,

$B=(100T)y\sin(2π×10^8t+kz)$

Wave number is given = K

frequency =$10^8Hz$

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

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

$E=cB=3×10^8×100y=3×10^{10}N/c$