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# Engineering

The magnetic field of a plane electromagnetic wave is given by :

\vec{B}=B_{0}\; \hat{i}\left [ cos\left ( kz-\omega t \right ) \right ]+B_{1}\hat{j}\; cos\left ( kz+\omega t \right )

where B_{0}=3\times 10^{-5}T and B_{1}=2\times 10^{-6}T.

The rms value of the force experienced by a stationary charge Q=10^{-4}C at Z=0 is closest to :

  • Option 1)

     0.6 \: N

  • Option 2)

    0.1\: N

  • Option 3)

    0.9\: N

  • Option 4)

    3\times 10^{-2}\; N

 

Answers (1)
S solutionqc

B=B_{0}\; cos\left ( kx-\omega t \right )\hat{i}+B_{1}\; cos \left ( kz+\omega t \right )\hat{j}

      B_{0}=3\times 10^{-5}T

       B_{1}=2\times 10^{-6}T

Q=10^{-4}C

F_{rms}=Q\sqrt{\left ( \frac{CB_{0}}{\sqrt{2}} \right )^{2}+\left ( \frac{CB_{1}}{\sqrt{2}} \right )^{2}}

           =10^{-4}\times \frac{3\times 10^{8}}{\sqrt{2}}\sqrt{\left ( 3\times 10^{-5} \right )^{2}+\left ( 2\times 10^{-6} \right )^{2}}

          

           =10^{-4}\times \frac{3\times 10^{8}}{\sqrt{2}}\sqrt{9\times 10^{-10}+4\times 10^{-12}}

             =10^{-4}\times \frac{3\times 10^{8}}{\sqrt{2}}\sqrt{900\times 10^{-12}+4\times 10^{-12}}

             =\frac{3\times 10^{4}}{\sqrt{2}}\sqrt{90{4}}\times 10^{-6}

          =0.63\; N

 

 

 

 

 


Option 1)

 0.6 \: N

Option 2)

0.1\: N

Option 3)

0.9\: N

Option 4)

3\times 10^{-2}\; N

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