Question 30: EM waves

Question

A plane electromagnetic wave is moving in free space with velocity $c = 3 	imes 10^8$ m/s and its electric field is given as $\vec{E}=54\sin(kz - \omega t)\,\hat{j}$ V/m, where $\hat{j}$ is the unit vector along y-axis. The magnetic field vector $\vec{B}$ of the wave is :

Accepted Answer

In an electromagnetic wave traveling through a vacuum, the amplitudes of the electric field ( $\mathrm{E}_0$ ) and the magnetic field ( $B_0$ ) are related by the speed of light ( $c$ ):
$$ c=\frac{E_0}{B_0} \Rightarrow B_0=\frac{E_0}{c} $$
Given that $\mathrm{E}_0=54 \mathrm{~V} / \mathrm{m}$ and $\mathrm{c}=3 	imes 10^8 \mathrm{~m} / \mathrm{s}$
$$ B_0=\frac{54}{3 	imes 10^8} $$
$$ \mathrm{B}_0=1.8 	imes 10^{-7} \mathrm{~T} $$
In a transverse electromagnetic wave, the electric field (E), magnetic field (B), and the direction of propagation (v) are all mutually perpendicular. The direction follows the right-hand rule:
$$ \widehat{\mathrm{E}} 	imes \widehat{\mathrm{B}}=\widehat{\mathrm{v}} $$
The variable $z$ in phase expression ( $k z-\omega t$ ) indicates the wave is moving along the $+z$ direction ( $\hat{k}$ ).
The unit vector along direction of electric field ( $\overrightarrow{\mathrm{E}}$ ) is given as $\hat{\mathrm{j}}$.
We need a unit vector such that $\hat{\mathrm{j}} 	imes(\widehat{\mathrm{B}})=\hat{\mathrm{k}}$.
We know that $\hat{\jmath} 	imes \hat{i}=-\hat{\mathrm{k}}$ and $-\hat{\jmath} 	imes \hat{i}=\hat{\mathrm{k}}$
Therefore, if $\widehat{\mathrm{E}}=\hat{\mathrm{j}}$ and $\widehat{\mathrm{v}}=\widehat{\mathrm{k}}$, then $\widehat{\mathrm{B}}$ must be $-\hat{\mathrm{i}}$ because $\hat{\mathrm{j}} 	imes(-\mathrm{i})=-(\hat{\mathrm{j}} 	imes \hat{\mathrm{i}})=-(-\widehat{\mathrm{k}})=\widehat{\mathrm{k}}$.
Combining the magnitude and the direction:
$$ \overrightarrow{\mathrm{B}}=-1.8 	imes 10^{-7} \sin (\mathrm{kz}-\omega \mathrm{t}) \hat{\mathrm{i}} \mathrm{~T} $$
Hence, the correct option is (A).

Answer

$-1.8	imes 10^{-7}\sin(kz - \omega t)\,\hat{i}$ T

Answer

$+1.8	imes 10^{-7}\sin(kz - \omega t)\,\hat{i}$ T

Answer

$1.4	imes 10^{-7}\sin(kz - \omega t)\,\hat{k}$ T

Answer

$1.4	imes 10^{-7}\sin(kz - \omega t)\,\hat{i}$ T

EM waves

A plane electromagnetic wave is moving in free space with velocity c = 3 \times 10^8 m/s and its electric field is given as \vec{E}=54\sin(kz - \omega t)\,\hat{j} V/m, where \hat{j} is the unit vector along y-axis. The magnetic field vector \vec{B} of the wave is :

Choose the correct answer: