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  • In a plane, electromagnetic wave the electric field oscillates sinusoidally at a frequency of <img alt="\mathrm{2 \times 10^{10} \mathrm{~Hz}}" src="https://entrancecorner.oncodecogs.com/gif.latex?

In a plane, electromagnetic wave the electric field oscillates sinusoidally at a frequency of \mathrm{2 \times 10^{10} \mathrm{~Hz}} and amplitude. \mathrm{48 \mathrm{Vm}^{-1}}. Then, which one of the following statement is true?

Option: 1

Wavelength of the wave is 2 \times 10^5 \mathrm{~m}


Option: 2

Amplitude of oscillating magnetic field is 48 T


Option: 3

Average energy density of electric field equals the average energy density of magnetic field


Option: 4

None of the above


Answers (1)

best_answer

Wavelength \mathrm{\lambda=\frac{c}{f}=\frac{3 \times 10^8}{2 \times 10^{10}}=1.5 \times 10^{-2} \mathrm{~m}}

Hence, option (1) is false

                  \begin{aligned} & \mathrm{B_0=\frac{E_0}{C}=\frac{48}{3 \times 10^8}} \\ \\& =16 \times 10^{-8} \mathrm{~T}=1.6 \times 10^{-7} \mathrm{~T} \end{aligned}

Energy density of electric field is 

                    \mathrm{U_C=\frac{1}{2} \varepsilon_0 E^2=\frac{1}{2} \varepsilon_0\left(\frac{E_0}{\sqrt{2}}\right)^2=\frac{1}{4} \varepsilon_0 E_0^2}

Energy density of magnetic field is 

                     \mathrm{U_B=\frac{1}{2 \mu_0} B^2=\frac{1}{2 \mu_0}\left(\frac{B_0}{\sqrt{2}}\right)^2}

                             \begin{aligned} & \mathrm{=\frac{1}{4 \mu} \cdot B_0^2=\frac{B_0^2}{4 \mu_0}} \quad \mathrm{\left[\therefore B_0=\frac{E_0}{c}\right] }\\ \\& =\mathrm{\frac{\left(\frac{E_0}{c}\right)^2}{4 \mu_0}=\frac{1}{4 \mu_0^2} \frac{E_0^2}{c^2}} \end{aligned}

                               \begin{aligned} & \mathrm{=\frac{E_0^2}{4 \mu_0\left(\frac{1}{\sqrt{\mu_0 \varepsilon_0}}\right)^2} \quad\left[\therefore c=\frac{1}{\sqrt{\mu_0 \varepsilon_0}}\right]} \\ \\& \mathrm{=\frac{\mu_0 \varepsilon_0}{4 \mu_0} E_0^2=\frac{1}{4} \varepsilon_0 E_0^2} \end{aligned}

Hence, option (3) is true.

Posted by

Kuldeep Maurya

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