Atraveling electromagnetic wave in a vacuum has an electric field amplitude of 69.1 v/m . calculate the intensity of this wave. then, determine the amount of energy that flows through area of 0.0247 m^2 over an interval of 13.1 s, assuming that the area is perpendicular to the direction of wave propagation.

The intensity of this wave and energy is 6.3385 N/m² and 2.0509 J.

Explanation:

Given that,

Electric field amplitude E₀= 69.1 V/m

Area A= 0.0247 m²

Time t= 13.1 s

We need to calculate the intensity

Using formula of intensity

Where, c = speed of light

Put the value into the formula

(b). We need to calculate the energy

Using formula of energy

Where, A = area

t = time

S = intensity

Put the value into the formula

Hence, The intensity of this wave and energy is 6.3385 N/m² and 2.0509 J.

answer: option c. i.e. electric current flowing “downhill” from the negative electrode to the positive electrode in a battery

explanation:

the water analogy to the circuits define the pipes (water flows) to the electric current. now, we know that the electric current flows from the positive terminal to the circuit (as shown in the diagram attached).

hence, option c is the correct answer.