Question 148: EM waves

Question

The electric field component of a monochromatic radiation is given by
$$\overrightarrow E $$ = 2 E0 $$\widehat i$$ cos kz cos $$\omega $$t
Its magnetic field $$\overrightarrow B $$  is then given by :

Accepted Answer

We have
$${{dE} \over {dz}} = {{ - dE} \over {dt}}$$
$${{dE} \over {dz}} =  - 2{E_0}k\sin kz\cos \omega t = {{ - dB} \over {dt}}$$
Therefore, $$dB =  + 2{E_0}k\sin kz\cos \omega t\,dt$$
That is, $$B =  + 20{E_0}k\sin {k_z}\cos \omega t\,dt$$
$$B =  + 2{E_0}k\sin {k_2}\int {\cos \omega t\,dt =  + 2{E_0}{k \over \omega }\sin kz\sin \omega t} $$
Now, $${{{E_0}} \over {{B_0}}} = {\omega  \over k} = c$$
Therefore, its magnetic field $$\overrightarrow B $$ is given by $$B = {{2{E_0}} \over c}\widehat j\sin kz\sin \omega t$$

Answer

$${{2{E_0}} \over c}$$ $$\widehat j$$ sin kz cos $$\omega $$t

Answer

$$-$$ $${{2{E_0}} \over c}$$ $$\widehat j$$ sin kz sin $$\omega $$t

Answer

$${{2{E_0}} \over c}$$ $$\widehat j$$ sin kz sin $$\omega $$t

Answer

$${{2{E_0}} \over c}$$ $$\widehat j$$ cos kz cos $$\omega $$t

EM waves

The electric field component of a monochromatic radiation is given by $\overrightarrow E = 2 E0 \widehat i cos kz cos \omega t Its magnetic field \overrightarrow B $ is then given by :

Choose the correct answer: