Question

The magnetic field of an electromagnetic wave is given by :
$\vec{B}=1.6\times 10^{-6}\cos \left(2\times 10^{7}z+6\times 10^{15}t\right)\left(2\hat{i}+\hat{j}\right)\frac{Wb}{m^2}$
The associated electric field will be :-

• A. $\vec{E}=4.8\times 10^2\cos \left(2\times 10^{7}z+6\times 10^{15}t\right)\left(\hat{i}-2\hat{j}\right)\frac{V}{m}$
• B. $\vec{E}=4.8\times 10^2\cos \left(2\times 10^{7}z-6\times 10^{15}t\right)\left(2\hat{i}+\hat{j}\right)\frac{V}{m}$
• C. $\vec{E}=4.8\times 10^2\cos \left(2\times 10^{7}z-6\times 10^{15}t\right)\left(-2\hat{i}+\hat{j}\right)\frac{V}{m}$
• D. $\vec{E}=4.8\times 10^2\cos \left(2\times 10^{7}z+6\times 10^{15}t\right)\left(-1\hat{i}+2\hat{j}\right)\frac{V}{m}$

Answer 1

Answer: (D)

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 $(-\hat{i}+2\hat{j})$