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  • A plane EM wave travelling in vacuum along z-direction is given by E=E_{0} sin (kz- omega t) hat{i} and B=B_{0} sin (kz-omega t)hat{j} a) evaluate oint E.dl over the rectangular loop 1234 shown in the figure

A plane EM wave traveling in vacuum along z-direction is given by

E=E_{0}\sin (kz-\omega t)\hat{i}\; and \; B=B_{0}\sin (kz-\omega t)\hat{j}

a) evaluate \oint E.dl over the rectangular loop 1234 shown in the figure

b) evaluate \int B.ds over the surface bounded by loop 1234

c) use an equation \oint E.dl=\frac{-d\phi _{B}}{dt} to prove \frac{E_{0}}{B_{0}}=c

d) by using a similar process and the equation

\oint B.dl=\mu _{0}I\epsilon _{0}\frac{-d\phi _{E}}{dt} prove that c=\frac{1}{\sqrt{\mu _{0}\epsilon _{0}}}

 

Answers (2)

 

 

Explanation:-

a) This part involves evaluating the line integral of the electric field E along a rectangular loop. This type of integral often arises in the context of Faraday’s Law of Induction,

 \oint \overrightarrow{E}.\overrightarrow{dl}=E_{0}h\left [ \sin (kz_{2}-\omega t)-\sin (kz_{1}-\omega t) \right ]

b)

This is a surface integral of the magnetic field B over the surface enclosed by the rectangular loop. This type of integral is related to Ampère's Law with Maxwell’s correction,

\oint \overrightarrow{B}.\overrightarrow{ds}=-\frac{B_{0}h}{k}\left [ \cos (kz_{2}-\omega t)-\cos (kz_{1}-\omega t) \right ]

c) Substituting the above equations in the following equation we get

\oint E.dl=-\frac{d\phi _{B}}{dt}=-\frac{d}{dt}\oint B.ds

\\E_{0}h\left [ \sin (kz_{2}-\omega t)-\sin (kz_{1}-\omega t) \right ]=\frac{B_0hw}{k} [ \sin (kz_{2}-\omega t)-\sin (kz_{1}-\omega t)]\\

 

\frac{E_{0}}{B_{0}}=C

d)

This is another integral form of Ampère’s Law with Maxwell’s correction, which accounts for both conduction currents and displacement currents. The left side of the equation is the line integral of the magnetic field around a closed loop.

\oint B.dl=\mu _{0}I\epsilon _{0}\frac{-d\phi _{E}}{dt}

\\B_0 h[\sin (kz_{2}-\omega t)-\sin (kz_{1}-\omega t)]=\frac{E_0h}{kw} [\sin (kz_{2}-\omega t)-\sin (kz_{1}-\omega t)]

\\B_0 h=\frac{E_0h}{kw}

We get

c=\frac{1}{\sqrt{\mu _{0}\varepsilon _{0}}}

Posted by

infoexpert23

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 [\sin (kz_{2}-\omega t)-\sin (kz_{1}-\omega t)] 

Posted by

Safeer PP

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