A plane electromagnetic wave of wavelength λ has an intensity I ....

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:

Options

A $\overset{\to}{E} = \sqrt{\dfrac{2I}{{ε}_{0}c}}\cos \left[\dfrac{2\pi}{\lambda}\left(y - ct\right)\right]\overset{^}{k}$ ,

$\overset{\to}{B} = + \dfrac{1}{c}E\overset{^}{i}$
B $\overset{\to}{E} = \sqrt{\dfrac{I}{{ε}_{0}c}}\cos \left[\dfrac{2\pi}{\lambda}\left(y - ct\right)\right]\overset{^}{k}$ ,

$\overset{\to}{B} = + \dfrac{1}{c}E\overset{^}{i}$
C $\overset{\to}{E} = \sqrt{\dfrac{2I}{{ε}_{0}c}}\cos \left[\dfrac{2\pi}{\lambda}\left(y + ct\right)\right]\overset{^}{k}$ ,

$\overset{\to}{B} = + \dfrac{1}{c}E\overset{^}{i}$
D $\overset{\to}{E} = \sqrt{\dfrac{I}{{ε}_{0}c}}\cos \left[\dfrac{2\pi}{\lambda}\left(y - ct\right)\right]\overset{^}{i}$ ,

$\overset{\to}{B} = + \dfrac{1}{c}E\overset{^}{k}$

Correct answer

A. $\overset{\to}{E} = \sqrt{\dfrac{2I}{{ε}_{0}c}}\cos \left[\dfrac{2\pi}{\lambda}\left(y - ct\right)\right]\overset{^}{k}$ ,

$\overset{\to}{B} = + \dfrac{1}{c}E\overset{^}{i}$

Step-by-step solution

{E}_{0}

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

I = \dfrac{1}{2}{ε}_{0}{E}_{0}^{2} \times c

. From the above relation, we get

{E}_{0} = \sqrt{\dfrac{2I}{c{ε}_{0}}}

. The relation between electric and magnetic fields is

{B}_{0} = \dfrac{{E}_{0}}{c}

. The direction of

\overset{\to}{E} \times \overset{\to}{B}

will be along

+ \overset{^}{j}

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