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# An electromagnetic wave passing through vacuum is described by the equations $E= { E }_{ 0 } \sin { \left( kx-\omega t \right) }$ and $B={ B }_{ 0 }\sin { \left( kx-\omega t \right) }$. Then,A. ${ E }_{ 0 }k={ B }_{ 0 }\omega$B. ${ E }_{ 0 }\omega={ B }_{ 0 }k$C. ${ E }_{ 0 }{ B }_{ 0 }=\omega k$D. ${ E }_{ 0 }={ B }_{ 0 }$

Last updated date: 18th Sep 2024
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Hint: Find the relation between electric field and magnetic field. They are related to each other by c i.e. speed of light, Write the equation for c in terms of wavelength. Then, get the expression for c in terms of angular frequency and wave number. Finally, substitute that value in the relation of electric field and magnetic field.

Given: Equations of electromagnetic waves: $E= { E }_{ 0 } \sin { \left( kx-\omega t \right) }$ and $B={ B }_{ 0 }\sin { \left( kx-\omega t \right) }$
The relation between Electric field ${ E }_{ 0 }$ and Magnetic field ${ B }_{ 0 }$ is given by,
${ E }_{ 0 }=c{ B }_{ 0 }$ …(1)
We know, $\nu =\dfrac { c }{ \lambda }$
Rearranging the above equation we get,
$c=\nu \lambda$ …(2)
We also know, $\nu =\dfrac { \omega}{ 2\pi }$
Thus, substituting this value in the equation. (2) we get,
$c=\dfrac { \omega}{ 2\pi }\lambda$
Now, substituting $\cfrac { 2\pi }{ \lambda } =k$ we get,
$c=\dfrac { \omega }{ k }$ …(2)
Then, substituting the equation. (3) in equation. (1) we get,
${ E }_{ 0 }=\dfrac { \omega }{ k } { B }_{ 0 }$
Rearranging the above equation we get,
${ E }_{ 0 }k={ B }_{ 0 }\omega$

So, the correct answer is “Option A”.

Note:
The direction of electromagnetic waves is found by vector cross product of the electric field and magnetic field. Electromagnetic radiations also show dual nature i.e. have the properties of a wave as well as the properties of a particle.