A point source of electromagnetic radiation has an average power output of $800 W$ . The maximum value of electric field at a distance $4.0 m$ from the source is :

64.7 V m^{- 1}

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57.8 V m^{- 1}

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56.72 V m^{- 1}

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$54.77 V m^{- 1}$

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Solution

The correct option is D
$54.77 V m^{- 1}$

Intensity of electromagnetic wave is
$I = \frac{P_{a v}}{2 π r^{2}} = \frac{E_{0}^{2}}{μ_{0} c}$
or
$E_{0} = \sqrt{\frac{μ_{0} c P_{a v}}{2 π r^{2}}}$
$μ_{0} = 4 π \times 10^{- 7}, P_{a v} = 800 W, r = 4 m, c = 3 \times 10^{8} m s^{- 2}) E_{0} = 54.77 V m^{- 1}$

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A point source of electromagnetic radiation has an average power output of 800 W. The maximum value of electric field at a distance 4.0 m from the source is :

The correct option is D 54.77 Vm−1 Intensity of electromagnetic wave is I=Pav2πr2=E20μ0c or E0=√μ0cPav2πr2 μ0=4π×10−7,Pav=800 W,r=4 m,c=3×108 ms−2)E0=54.77 Vm−1

A point source of electromagnetic radiation has an average power output of $800 W$ . The maximum value of electric field at a distance $4.0 m$ from the source is :