Browse by Stream
QnA
Home
QnA
Go To Your Study Dashboard

Get Answers to all your Questions

header-bg qa

The electric field component of a monochromatic radiation is given by


\underset{E}{\rightarrow} = 2E_{0}\: \: \hat{i}\: \: \cos kz\: \cos wt

Its magnetic field   \underset{B}{\rightarrow}is then given by :

 

  • Option 1)

    \frac{2E_{0}}{C}\: \hat{J}\sin kz\cos wt

  • Option 2)

    -\frac{2E_{0}}{c}\: \hat{j}\sin kz\sin wt

  • Option 3)

    \frac{2E_{0}}{c}\: \hat{j}\sin kz\sin wt

  • Option 4)

    \frac{2E_{0}}{c}\: \hat{j}\cos kz \cos wt

 

Answers (1)

best_answer

As we have learned

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 -

Combination of mutually perpendicular electric and magnetic field is referred to as Electromagnetic Wave.

-

 

 B_0 = \frac{2E_0 }{C }

It will be along Y 

 \vec B = \frac{2E_0 }{C }\cos K Z \cdot \cos wt

 

 

 

 

 

 

 

 


Option 1)

\frac{2E_{0}}{C}\: \hat{J}\sin kz\cos wt

Option 2)

-\frac{2E_{0}}{c}\: \hat{j}\sin kz\sin wt

Option 3)

\frac{2E_{0}}{c}\: \hat{j}\sin kz\sin wt

Option 4)

\frac{2E_{0}}{c}\: \hat{j}\cos kz \cos wt

Posted by

Avinash

View full answer

Similar Questions

Trending Articles/News

anam-khan

BITSAT 2024 application correction facility begins tomorrow; editable fields
anam-khan

BITSAT 2024 registration window closes tomorrow; exam from May 20
anam-khan

BITSAT 2024 registration deadline extended till April 16; session 1 exam from May 19 to 24

Latest Question