JEE Mains · Physics · STD 12 - 8. Electromagnetic waves
Magnetic field in a plane electromagnetic wave is given by \(\vec B = {B_0}\,\sin \,\left( {kx + \omega t} \right)\hat jT\) Expression for corresponding electric field will be Where \(c\) is speed of light
- A \(\vec E = {B_0}\,c\sin \,\left( {kx + \omega t} \right)\hat k\,V/m\)
- B \(\vec E = \frac{{{B_0}}}{c}\,\sin \,\left( {kx + \omega t} \right)\hat k\,V/m\)
- C \(\vec E = - {B_0}\,c\sin \,\left( {kx + \omega t} \right)\hat k\,V/m\)
- D \(\vec E = {B_0}\,c\sin \,\left( {kx - \omega t} \right)\hat k\,V/m\)
Answer & Solution
Correct Answer
(A) \(\vec E = {B_0}\,c\sin \,\left( {kx + \omega t} \right)\hat k\,V/m\)
Step-by-step Solution
Detailed explanation
Speed of \(EM\) wave in force space \(\left( c \right) = \frac{{{E_0}}}{{{B_0}}}\,\) or \(\vec E = c{B_0}{\mkern 1mu} \sin {\mkern 1mu} \left( {kx + \omega t} \right)\hat k\)
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