Final answer:
The energy delivered by a time-harmonic electromagnetic wave through a rectangular aperture in one hour is calculated by finding the energy flux using the Poynting vector, then multiplying by the area of the aperture and the time in seconds.
Step-by-step explanation:
To calculate the energy delivered by the electromagnetic wave to the other side of the screen in one hour, we first need to understand the basic properties of an electromagnetic wave and how it transfers energy. The energy flux of an electromagnetic wave can be described by the Poynting vector, which in free space is given by the formula S = (1/2)*(c*Em2/Z0), where c is the speed of light in vacuum (~3×108 m/s) and Z0 is the impedance of free space (~377 ohms). For an electric field with an amplitude Em = 5 V/m, we can calculate the energy flux and then multiply it by the area of the aperture and the time duration to find the total energy.
Using the given values, the energy flux S becomes S = (1/2)*(3×108 m/s)(5 V/m)2/377 ohms, which yields an energy flux in W/m2. The area A of the aperture can be calculated using the provided dimensions, a = 35 cm and b = (10+ N) cm. Converting these lengths to meters and multiplying them gives us the area in m2. Multiplying the energy flux by the area and by the number of seconds in an hour (3600 s) gives us the total energy transferred in one hour.