### JEE Main On 16 April 2018 Question 30

##### Question: A plane electromagnetic wave of wavelength $ \lambda $ has an intensity $ I. $ It is propagating along the positive $ Y-direction. $ The allowed expressions for the electric and magnetic fields are given by [JEE Main 16-4-2018]

#### Options:

A) $ \vec{E}=\sqrt{\frac{I}{{\varepsilon_0}C}}\cos [ \frac{2\pi }{\lambda }(y-ct) ]\hat{i};\vec{B}=\frac{1}{c}E\hat{k} $

B) $ \vec{E}=\sqrt{\frac{I}{{\varepsilon_0}C}}\cos [ \frac{2\pi }{\lambda }(y-ct) ]\hat{k};\vec{B}=-\frac{1}{c}E\hat{i} $

C) $ \vec{E}=\sqrt{\frac{2I}{{\varepsilon_0}C}}\cos [ \frac{2\pi }{\lambda }(y-ct) ]\hat{k};\vec{B}=+\frac{1}{c}E\hat{i} $

D) $ \vec{E}=\sqrt{\frac{2I}{{\varepsilon_0}C}}\cos [ \frac{2\pi }{\lambda }(y+ct) ]\hat{k};\vec{B}=\frac{1}{c}E\hat{i} $

## Show Answer

#### Answer:

Correct Answer: C

#### Solution:

E is the electric field vector, and B is the magnetic field vector of the EM wave.

For electromagnetic waves E and B are always perpendicular to each other and perpendicular to the direction of propagation.

The direction of propagation is the direction of E x B.

So, if the wave propagates in the +Y direction then the direction of E and B should be in +X and +Z or vice versa i.e +Z and +X respectively.

Case 1. Let us suppose $ \vec{E} $ is in $ \hat{i} $ and $ \hat{B} $ is in $ \hat{k} $ Then $ \vec{E}\times \vec{B} $ will be in $ -\hat{j} $ Not Possible.

Case 2. Let us suppose $ \vec{E} $ is in $ \hat{k} $ and $ \vec{B} $ is in $ \hat{i} $ Then $ \vec{E}\times \vec{B} $ will be in $ \hat{j} $

This is satisfying option C as the electric and magnetic field also propagate in positive y direction with time

so $ (y-ct) $ should be there in wave equation.

Also $ I=\frac{c{\in_o}}{2}|E _{o}{{|}^{2}} $ $ |E _{o}|=\sqrt{\frac{2I}{c{\in_o}}} $ From these, we can say that option C would be the best option.