Question

A pulse of radiation propagates with velocity $<0,0,-c>$. The electric field in the pulse is $<7.2\times {10}^6,0,0>\mathbf{}\mathit{N}\mathbf{/}\mathit{C}$. What is the magnetic field in the pulse?

Short answer

The magnetic field in the pulse is $<0,-0.024,0>\text{T}$.

Step-by-step solution

Given data

Velocity of the pulse:

Electric field of the pulse:

Magnetic field in a pulse

Magnitude of magnetic field in a pulse with magnitude of electric field is:

$\mathit{B}\mathbf{=}\frac{\mathbf{E}}{\mathbf{c}}$ …… (i)

Here, is the speed of light in vacuum with value $\mathit{c}\mathbf{=}\mathbf{3}\mathbf{\times }{\mathbf{10}}^{\mathbf{8}}\mathbf{}\mathit{m}\mathbf{/}\mathit{s}$$c=3\times {10}^{8}m/s$.

The direction of the magnetic field is such that the direction of the wave is given by the direction as follows:

$\overrightarrow{\mathbf{E}}\mathbf{\times }\overrightarrow{\mathbf{B}}$

Determine the magnetic field in the given pulse

From equation (i), the magnitude of the electric field is:

$$\begin{gathered} B=\frac{7.2\times {10}^{6}N/C}{3\times {10}^{8}m/s} \\ =0.024·\left(1N.s/C.m\times \frac{1T}{1N.s/C.m}\right) \\ =0.024\text{T} \end{gathered}$$

The direction of the wave is along $-z$axis and the direction of the electric field is along x axis. Hence the direction of the magnetic field has to be along $-y$axis.

Thus, the magnetic field is .