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Question

(A) The direction of propagation is given by $\hat{n} = \frac{\hat{i}+\hat{j}}{\sqrt{2}}$.The electric field polarization is given by $\hat{E} = \hat{

Vedclass · Accepted Answer

$B_{0} \frac{\hat{i}-\hat{j}}{\sqrt{2}} \cos \left(\omega t - k \frac{\hat{i}+\hat{j}}{\sqrt{2}} \cdot \vec{r}\right)$

Vedclass · Answer

(A) The direction of propagation is given by $\hat{n} = \frac{\hat{i}+\hat{j}}{\sqrt{2}}$.The electric field polarization is given by $\hat{E} = \hat{k}$.In an electromagnetic wave,the direction of propagation $\hat{n}$ is given by the cross product of the electric field direction $\hat{E}$ and the magnetic field direction $\hat{B}$,i.e.,$\hat{n} = \hat{E} 	imes \hat{B}$.Substituting the known values: $\frac{\hat{i}+\hat{j}}{\sqrt{2}} = \hat{k} 	imes \hat{B}$.We know that $\hat{k} 	imes \hat{i} = \hat{j}$ and $\hat{k} 	imes \hat{j} = -\hat{i}$.Therefore,$\hat{k} 	imes \left( \frac{\hat{i}-\hat{j}}{\sqrt{2}} ight) = \frac{\hat{j} - (-\hat{i})}{\sqrt{2}} = \frac{\hat{i}+\hat{j}}{\sqrt{2}}$.Thus,the direction of the magnetic field is $\hat{B} = \frac{\hat{i}-\hat{j}}{\sqrt{2}}$.Hence,the magnetic field vector is $\vec{B} = B_{0} \frac{\hat{i}-\hat{j}}{\sqrt{2}} \cos \left(\omega t - \vec{k} \cdot \vec{r}ight)$.

$A$ plane electromagnetic wave is propagating along the direction $\frac{\hat{i}+\hat{j}}{\sqrt{2}},$ with its polarization along the direction $\hat{k}$. The correct form of the magnetic field of the wave would be (here $B_{0}$ is an appropriate constant)

Similar Questions

The electric field $(E)$ and magnetic field $(B)$ of an electromagnetic wave passing through vacuum are given by
$E = E_0 \sin (kx - \omega t)$
$B = B_0 \sin (kx - \omega t)$
Then the correct statement among the following is

If the magnetic field of a plane electromagnetic wave is given by (The speed of light $c = 3 \times 10^8 \, m/s$):
$B = 100 \times 10^{-6} \sin \left[ 2\pi \times 2 \times 10^{15} \left( t - \frac{x}{c} \right) \right]$
Then the maximum electric field associated with it is:

$A$ sky wave with a frequency $55 \, MHz$ is incident on the $D$-region of the Earth's atmosphere at $45^\circ$. The angle of refraction is ......... $^\circ$ (electron density for the $D$-region is $400 \, \text{electrons}/cm^3$).

$A$ plane electromagnetic wave of frequency $50 \ MHz$ travels in free space. If the average energy densities in the electric field and magnetic field are $K_{E}$ and $K_{B}$ respectively,then the correct option in the following is

If $10 \%$ of the power of a $(100 \pi) \ W$ light bulb is converted to visible radiation,then the average intensity of visible radiation at a distance of $10 \ m$ is (in $W \ m^{-2}$)

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