Question

The expression of the electric field associated with an electromagnetic wave in vacuum is given by

E vector=(100vm^-1)z^sin(2π×10^8t+kx)

Determine the wave number, frequency, the direction of propagation and the magnitude and direction of the magnetic field associated with the wave.

Answer 1

a) The wave number can be found from the formula:

$$k=\dfrac{\omega}{v}=\dfrac{2\pi\cdot10^8\ \dfrac{rad}{s}}{3\cdot10^8\ \dfrac{m}{s}}=2.1$$

b) The frequency can be found as follows:

$$f=\dfrac{\omega}{2\pi}=\dfrac{2\pi\cdot10^8\ \dfrac{rad}{s}}{2\pi}=10^8\ Hz.$$

c) As we can see from the expression of the electric field, the direction of propagation is along the axis $x$.

d) From the relationship between the electric field, magnetic field and the speed of light, we have:

$$c=\dfrac{E}{B},$$ $$B=\dfrac{E}{c}=\dfrac{100\ \dfrac{V}{m}\cdot \hat{z}}{3\cdot10^8\ \dfrac{m}{s}}=3.3\cdot10^{-7}\ T\cdot\hat{z}.$$

The magnitude of the magnetic field associated with the wave is $3.3\cdot10^{-7}\ T$, the direction is along the axis $z$.