The electric field of an electromagnetic wave | vedclass

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Question

(D) The given electric field is $\overrightarrow{E} = E_0 \cos(\omega t - \vec{k} \cdot \vec{r}) \hat{n}_E$,where $\hat{n}_E = \frac{4\hat{i} - 3\hat{

Vedclass · Accepted Answer

$\overrightarrow{B}=-\frac{57}{3 	imes 10^8} \cos \left[7.5 	imes 10^6 t-5 	imes 10^{-3}(3 x+4 y)ight] (\hat{k})$

Vedclass · Answer

(D) The given electric field is $\overrightarrow{E} = E_0 \cos(\omega t - \vec{k} \cdot \vec{r}) \hat{n}_E$,where $\hat{n}_E = \frac{4\hat{i} - 3\hat{j}}{5}$ is the unit vector in the direction of the electric field.The wave vector is $\vec{k} = 5 	imes 10^{-3} (3\hat{i} + 4\hat{j}) = 1.5 	imes 10^{-2} \hat{i} + 2 	imes 10^{-2} \hat{j}$.The unit vector in the direction of propagation is $\hat{k} = \frac{3\hat{i} + 4\hat{j}}{5}$.The magnetic field direction is given by $\hat{n}_B = \hat{k} 	imes \hat{n}_E$.$\hat{n}_B = \left( \frac{3\hat{i} + 4\hat{j}}{5} ight) 	imes \left( \frac{4\hat{i} - 3\hat{j}}{5} ight) = \frac{1}{25} [3\hat{i} 	imes (-3\hat{j}) + 4\hat{j} 	imes 4\hat{i}] = \frac{1}{25} [-9\hat{k} - 16\hat{k}] = -\frac{25}{25} \hat{k} = -\hat{k}$.The magnitude of the magnetic field is $B_0 = \frac{E_0}{c} = \frac{57}{3 	imes 10^8}$.Thus,$\overrightarrow{B} = -\frac{57}{3 	imes 10^8} \cos \left[7.5 	imes 10^6 t - 5 	imes 10^{-3}(3 x + 4 y)ight] \hat{k}$.

The electric field of an electromagnetic wave in free space is $\overrightarrow{E}=57 \cos \left[7.5 \times 10^6 t-5 \times 10^{-3}(3 x+4 y)\right]\ (4 \hat{i}-3 \hat{j})\ N/C$. The associated magnetic field in Tesla is

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