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# Give answer! - Electromagnetic Waves - JEE Main-2

The electric field of a plane electromagnetic wave is given by

$\vec{E}=E_{o}\hat{i}cos(kz)cos(\omega t)$

The corresponding magnetic field $\vec{B}$ is then given by :

• Option 1)

$\vec{B}=\frac{E_{o}}{C}\: \hat{j}\: sin(kz)sin(\omega t)$

• Option 2)

$\vec{B}=\frac{E_{o}}{C}\: \hat{j}\: sin(kz)cos(\omega t)$

• Option 3)

$\vec{B}=\frac{E_{o}}{C}\: \hat{j}\: cos(kz)sin(\omega t)$

• Option 4)

$\vec{B}=\frac{E_{o}}{C}\: \hat{k}\: sin(kz)cos(\omega t)$

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Relation between Eo and Bo -

$E_{o}=c.B_{o}$

- wherein

$E_{o}$= Electric field amplitude

$B_{o}$ = Magnetic field amplitude

C= Speed of light in vacuum

$\vec{E}=E_{o}\hat{i}cos(kz)cos(\omega t)$

To know $\vec{B}$

$\frac{dE}{dz}=-\frac{dB}{dt}$

$-E_o(cos\omega t)(-sinkz).k=-\frac{dB}{dt}$

$B=E_o(cos\omega t)(-sinkz).kdt$

$B=E_ok\frac{(sin\omega t)}{\omega }(sinkz)$

$B=\frac{E_o}{C}(sinkz)(sin\omega t)$

Now,

$\vec{E}\times \vec{B}=$ Gives direction of wave

As wave is along z ($\hat{k}$)

$\vec{E}$ is along $\hat{i}$

So, $\vec{B}$ is along $\hat{j}$

So,

$B=\frac{E_o}{C}(sinkz)(sin\omega t)\hat{j}$

Option 1)

$\vec{B}=\frac{E_{o}}{C}\: \hat{j}\: sin(kz)sin(\omega t)$

Option 2)

$\vec{B}=\frac{E_{o}}{C}\: \hat{j}\: sin(kz)cos(\omega t)$

Option 3)

$\vec{B}=\frac{E_{o}}{C}\: \hat{j}\: cos(kz)sin(\omega t)$

Option 4)

$\vec{B}=\frac{E_{o}}{C}\: \hat{k}\: sin(kz)cos(\omega t)$

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