question_answer

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 :-             [JEE Main 8-4-2019 Afternoon]

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) \[\vec{E}=4.8\times {{10}^{2}}\cos \left( 2\times {{10}^{7}}z-6\times {{10}^{15}}t \right)\left( -2\hat{j}+\hat{i} \right)\frac{V}{m}\]

D) \[\vec{E}=4.8\times {{10}^{2}}\cos \left( 2\times {{10}^{7}}z+6\times {{10}^{15}}t \right)\left( -\hat{i}+2\hat{j} \right)\frac{V}{m}\]

Correct Answer: A

Solution :

[a] If we use that direction of light propagation will be along  Then [d] option is correct. Detailed solution is as following.

magnitude of E = CB

and  are perpendicular to each other

either direction of  is  or from given option

Also wave propagation direction is parallel to

which is

is along