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  • The ratio of contributions made by the electric field and magnetic field components to the intensity of an EM wave is:  Option: 1 <img al

The ratio of contributions made by the electric field and magnetic field components to the intensity of an EM wave is:
 

Option: 1

\mathrm{c: 1}

 


Option: 2

\mathrm{\mathrm{c}^2: 1}
 


Option: 3

\mathrm{1: 1}
 


Option: 4

\sqrt{\mathrm{c}}: 1


Answers (1)

best_answer

Intensity in terms of electric field, \mathrm{\mathrm{u}_{\mathrm{av}}=\frac{1}{2} \varepsilon_0 \mathrm{E}_0^2}

Intensity in terms of magnetic field, \mathrm{\mathrm{u}_{\mathrm{av}}=\frac{1}{2} \frac{\mathrm{B}_0^2}{\mu_0}}

Now, taking the intensity in terms of electric field.

\mathrm{\left(\mathrm{U}_{\mathrm{av}}\right)_{\text {electric field }}=\frac{1}{2} \varepsilon_0 \mathrm{E}_0^2 }

\mathrm{\Rightarrow=\frac{1}{2} \varepsilon_0\left(\mathrm{cB}_0\right)^2 \quad \quad {\left[\because \mathrm{E}_0=\mathrm{cB}_0\right]} }

\mathrm{=\frac{1}{2} \varepsilon_0 \times \mathrm{c}^2 \mathrm{~B}_0^2 }

But,\mathrm{ \quad \mathrm{c}=\frac{1}{\sqrt{\mu_0 \varepsilon_0}} }

\mathrm{ \therefore \quad\left(\mathrm{u}_{\mathrm{av}}\right)_{\text {electric field }}=\frac{1}{2} \varepsilon_0 \times \frac{1}{\mu_0 \varepsilon_0} \mathrm{~B}_0^2=\frac{1}{2} \frac{\mathrm{B}_0^2}{\mu_0} \\ }

\mathrm{=\left(\mathrm{u}_{\mathrm{av}}\right)_{\text {magnetic field }}}

Thus, the energy in electromagnetic wave is divided equally between electric field vector and magnetic field vector. Therefore, the ratio of contributions by the electric field and magnetic field components to the intensity of an electromagnetic wave is 1 : 1.

Hence option 3 is correct.


 

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