Practicing Success

Target Exam

CUET

Subject

Physics

Chapter

Electromagnetic Waves

Question:

A point source of electromagnetic radiation has an average power output of 800W. The maximum value of electric field at a distance 3.5m from the source will be:

Options:

56.7 V/m

62.6 V/m

39.3 V/m

47.5 V/m

Correct Answer:

62.6 V/m

Explanation:

Intensity of electromagnetic wave given is by

I = $\frac{P_{a v}}{4 \pi r^2}=\frac{E^2 m}{2 \mu_0 c}$

$E_m=\sqrt{\frac{\mu_0 c P_{a v}}{2 \pi r^2}}$

$=\sqrt{\frac{\left(4 \pi \times 10^{-7}\right) \times\left(3 \times 10^8\right) \times 800}{2 \pi \times 3.5^2}}$ = 62.6 V/m