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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)

 
Answers (1)
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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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