Question

The magnetic field of an electromagnetic wave is given by : $\vec{B} =1.6 \times10^{-6} \cos\left(2\times10^{7} z +6 \times10^{15}t\right)\left(2\hat{i} +\hat{j}\right) \frac{Wb}{m^{2}}$ The associated electric field will be :-

• A. $\vec{E} =4.8 \times10^{2} \cos\left(2\times10^{7}z + 6\times10^{15}t\right)\left(\hat{i} -2\hat{j}\right) \frac{V}{m}$
• B. $\vec{E} =4.8 \times10^{2} \cos\left(2\times10^{7}z - 6\times10^{15}t\right)\left(2\hat{i} + \hat{j}\right) \frac{V}{m}$
• C. $\vec{E} =4.8 \times10^{2} \cos\left(2\times10^{7}z - 6\times10^{15}t\right)\left( - 2\hat{i} + \hat{j}\right) \frac{V}{m}$
• D. $\vec{E} =4.8 \times10^{2} \cos\left(2\times10^{7}z + 6\times10^{15}t\right)\left( -1 \hat{i} + 2\hat{j}\right) \frac{V}{m}$

Answer 1

Answer: (A)

$\vec{E} =4.8 \times10^{2} \cos\left(2\times10^{7}z + 6\times10^{15}t\right)\left(\hat{i} -2\hat{j}\right) \frac{V}{m}$

Answer 2

If we use that direction of light propagation will be along $\vec{E} \times \vec{B}$? Then (4) option is correct.
Detailed solution is as following.
magnitude of E = CB
$E = 3 \times 10^8 \times 1.6 \times 10^{-6} \times \sqrt{5}$
$E = 4.8 \times 10^2 \; \sqrt{5}$
$\vec{E}$ and $\vec{B}$ are perpendicular to each other
$\Rightarrow \; \vec{E} . \vec{B} = 0$
$\Rightarrow$ either direction of $\vec{E}$ is $\hat{i} - 2\hat{j}$ or $-\hat{i} + 2 \hat{j}$ from given option
Also wave propagation direction is parallel to
$\vec{E} \times \vec{B}$ which is $- \hat{k}$
$\Rightarrow \; \vec{E}$ is along $( - \hat{i} + 2 \hat{j})$