Get Answers to all your Questions

header-bg qa

A plane electromagnetic wave having a frequency \nu =23.9 GHz propagates along the positive z-direction in free space. The peak value of the Electric Field is 60 V/m. Which among the following is the acceptable magnetic field component in the electromagnetic wave?

 

  • Option 1)

    \vec{B}=2\times 10^{7}\sin (0.5\times 10^{3}z+1.5\times 10^{11}t)\hat{i}

     

     

     

  • Option 2)

    \vec{B}=2\times 10^{-7}\sin (0.5\times 10^{3}z+1.5\times 10^{11}t)\hat{i}

  • Option 3)

    \vec{B}=60\sin (0.5\times 10^{3}x+1.5\times 10^{11}t)\hat{k}

  • Option 4)

    \vec{B}=2\times 10^{-7}\sin (1.5\times 10^{2}x+0.5\times 10^{11}t)\hat{j}

Answers (1)

best_answer

 

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

 

 

Electromagnetic wave propagate in =z-direction

Let B=B_{0}\sin (kz-wt)

B=B_{0}=\frac{E}{C}=\frac{60}{3\times 10^{8}}=2\times 10^{^{-7}}T

k=0.5\times 10^{3}

w=1.5\times 10^{11}

so correct option is 

\vec{B}=2\times 10^{-7}\sin (0.5\times 10^{3}z+1.5\times 10^{11}t)\hat{i}


Option 1)

\vec{B}=2\times 10^{7}\sin (0.5\times 10^{3}z+1.5\times 10^{11}t)\hat{i}

 

 

 

Option 2)

\vec{B}=2\times 10^{-7}\sin (0.5\times 10^{3}z+1.5\times 10^{11}t)\hat{i}

Option 3)

\vec{B}=60\sin (0.5\times 10^{3}x+1.5\times 10^{11}t)\hat{k}

Option 4)

\vec{B}=2\times 10^{-7}\sin (1.5\times 10^{2}x+0.5\times 10^{11}t)\hat{j}

Posted by

solutionqc

View full answer

JEE Main high-scoring chapters and topics

Study 40% syllabus and score up to 100% marks in JEE