Part A The intensity of an electromagnetic wave is 30 W/m2 What will the intensity be if the amplitude of the electric field is doubled? Express your answer with the appropriate units. Subm Previous Answers Completed; correct answer withheld by Instructor – Part B What will the intensity be if the amplitude of the magnetic field is doubled? Express your answer with the appropriate units. 1? I = Value Units Submit Request Answer Part What will the intensity be if the amplitudes of both the electric and the magnetic fields are doubled? Express your answer with the appropriate units. (s17 I = Value Units Submit Request Answer Part What will the intensity be if the frequency is doubled? Express your answer with the appropriate units. 1924 I = Value Units Submit Request Answer
The correct answer and explanation is:
To solve this, let’s break the problem into parts while considering the relationships between the intensity II, electric field amplitude E0E_0, and magnetic field amplitude B0B_0.
Part A: Doubling the electric field amplitude
The intensity II of an electromagnetic wave is proportional to the square of the amplitude of the electric field (E0E_0): I∝E02I \propto E_0^2
If the electric field amplitude E0E_0 is doubled (E0′=2E0E_0′ = 2E_0), the intensity becomes: I′=(2E0)2∝4E02I’ = (2E_0)^2 \propto 4E_0^2
This means the intensity will increase by a factor of 4. Therefore: I′=4×30 W/m2=120 W/m2I’ = 4 \times 30 \, \text{W/m}^2 = 120 \, \text{W/m}^2
Part B: Doubling the magnetic field amplitude
The intensity II is also proportional to the square of the magnetic field amplitude (B0B_0): I∝B02I \propto B_0^2
If the magnetic field amplitude B0B_0 is doubled (B0′=2B0B_0′ = 2B_0), the intensity becomes: I′=(2B0)2∝4B02I’ = (2B_0)^2 \propto 4B_0^2
Thus, the intensity will again increase by a factor of 4: I′=4×30 W/m2=120 W/m2I’ = 4 \times 30 \, \text{W/m}^2 = 120 \, \text{W/m}^2
Part C: Doubling both electric and magnetic field amplitudes
The intensity depends on both the electric and magnetic fields through the energy density relationship: I∝E02 and B02I \propto E_0^2 \, \text{and} \, B_0^2
Doubling both E0E_0 and B0B_0 means: I′=(2E0)2×(2B0)2=4E02×4B02=16×II’ = (2E_0)^2 \times (2B_0)^2 = 4E_0^2 \times 4B_0^2 = 16 \times I
Hence, the new intensity is: I′=16×30 W/m2=480 W/m2I’ = 16 \times 30 \, \text{W/m}^2 = 480 \, \text{W/m}^2
Part D: Doubling the frequency
The intensity II is independent of the frequency of the wave. While the frequency affects the wavelength and energy per photon (E=hfE = h f), the intensity depends only on the amplitudes of the electric and magnetic fields, not their frequency. Therefore: I′=I=30 W/m2I’ = I = 30 \, \text{W/m}^2
Summary of Results:
- Part A: I=120 W/m2I = 120 \, \text{W/m}^2
- Part B: I=120 W/m2I = 120 \, \text{W/m}^2
- Part C: I=480 W/m2I = 480 \, \text{W/m}^2
- Part D: I=30 W/m2I = 30 \, \text{W/m}^2
These results rely on the proportional relationship of intensity to the square of the field amplitudes and the fact that intensity is independent of wave frequency.