AP EAMCET201922 Apr 2019Morning ShiftPhysicsDual Nature of MatterActual
A point source of electromagnetic radiation has an average power output of (960 ~W ). The peak value of the electric field at a distance (400 ~cm ) from the source is
Options
- A(60 Vm ⁻¹ )
- B(120 Vm ⁻¹ )
- C(30 Vm ⁻¹ )
- D(180 Vm ⁻¹ )
Correct answer
A. (60 Vm ⁻¹ )
Step-by-step solution
Given, average power output, (P=960 ~W ) Distance, (r=400 ~cm =4 ~m ) Intensity of (E M ) waves is given by ( aligned I & = P 4 r^2 = 1 2 ₀ E₀^2 c E₀^2= P 2 r^2 ₀ c E & = P 2 r^2 ₀ c & = 960 2 3.14 4^2 8.85 10⁻¹² 3 10^8 & = 0.36 10^4 =0.6 10^2=60 Vm ⁻¹ aligned )