Get Answers to all your Questions

header-bg qa

Radio station on the surface of earth radiates 50kw. If transmitter radiates equally in all directions above the surface of earth, What is the amplitude of electric field detected 100km away

Option: 1

2.45vm^{-1}


Option: 2

0.245vm^{-1}


Option: 3

0.0245vm^{-1}


Option: 4

0.00245vm^{-1}


Answers (1)

best_answer

As we have learnt in

 

Property of EM wave -

Its propagation obeys the inverse square law.

-

 

 

I=\frac{P}{2\pi r^{2}}\Rightarrow I\propto \frac{1}{\gamma ^{2}}

I=\frac{1}{2}\epsilon oE^{2}c,\, \, \frac{1}{2}E_{o}^{2}c=\frac{P}{2\pi r^{2}}

E_{o}=\sqrt{\frac{P}{\epsilon oc\pi r^{2}}}

E_{o}=\sqrt{\frac{5\times 10^{3}}{\left ( 8.85\times 10^{-12} \right )\times \left ( 3\times 10^{8} \right )(3.14)\times (10^{5})^{2}}}

E_{o}=0.0245vm^{-1}

Posted by

shivangi.bhatnagar

View full answer

JEE Main high-scoring chapters and topics

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