Q73 / 90 · JEE Main 2018 (16 Apr Online) · Physics

A plane electromagnetic wave of wavelength $λ$ has an intensity $I$ . It is propagating along the positive $Y -$ direction. The allowed expressions for the electric and magnetic fields are given by:

Correct answer
A
$\vec{E} = \sqrt{\frac{2 I}{ε_{0} c}} \cos [\frac{2 π}{λ} (y - c t)] \hat{k}$ ,

$\vec{B} = + \frac{1}{c} E \hat{i}$
B
$\vec{E} = \sqrt{\frac{I}{ε_{0} c}} \cos [\frac{2 π}{λ} (y - c t)] \hat{k}$ ,

$\vec{B} = + \frac{1}{c} E \hat{i}$
C
$\vec{E} = \sqrt{\frac{2 I}{ε_{0} c}} \cos [\frac{2 π}{λ} (y + c t)] \hat{k}$ ,

$\vec{B} = + \frac{1}{c} E \hat{i}$
D
$\vec{E} = \sqrt{\frac{I}{ε_{0} c}} \cos [\frac{2 π}{λ} (y - c t)] \hat{i}$ ,

$\vec{B} = + \frac{1}{c} E \hat{k}$

Solution

E_{0}

is the amplitude of the electric field then the intensity of the electromagnetic wave is given as

I = \frac{1}{2} ε_{0} E_{0}^{2} \times c

. From the above relation, we get

E_{0} = \sqrt{\frac{2 I}{c ε_{0}}}

. The relation between electric and magnetic fields is

B_{0} = \frac{E_{0}}{c}

. The direction of

\vec{E} \times \vec{B}

will be along

+ \hat{j}

Hence, the correct answer is option: 'A'.