Math Problem Statement

A plane wave travelling in a non-magnetic medium has electric field intensity E=50sin(108t-2z)ak kV/m. Calculate the average power in kW crossing a circular patch of diameter 12 cm in z constant plane.

Solution

To calculate the average power crossing a circular patch in a non-magnetic medium, we can follow these steps:

Step 1: Given Data

The electric field intensity is given by: E=50sin(108t2z)a^kkV/mE = 50 \sin(10^8 t - 2z) \, \hat{a}_k \, \text{kV/m} This is a plane wave propagating in the z-direction with an amplitude of 50 kV/m.

Step 2: Key Equations

In an electromagnetic wave, the average power crossing a surface is given by the time-averaged Poynting vector S\langle \mathbf{S} \rangle, which represents power per unit area. The Poynting vector is expressed as: S=1μ0(E×H)\mathbf{S} = \frac{1}{\mu_0} (\mathbf{E} \times \mathbf{H}) However, for the average power, we use: S=12ηE02\langle S \rangle = \frac{1}{2\eta} |E_0|^2 where:

  • E0E_0 is the amplitude of the electric field.
  • η\eta is the intrinsic impedance of the medium.
  • μ0=4π×107H/m\mu_0 = 4\pi \times 10^{-7} \, \text{H/m} is the permeability of free space.

For a non-magnetic medium, the intrinsic impedance η\eta is approximately the intrinsic impedance of free space: η0=377Ω\eta_0 = 377 \, \Omega

Step 3: Electric Field Amplitude

The amplitude of the electric field is given as E0=50kV/m=50×103V/mE_0 = 50 \, \text{kV/m} = 50 \times 10^3 \, \text{V/m}.

Step 4: Area of the Circular Patch

The patch is circular with a diameter of 12 cm, so the radius rr is: r=122=6cm=0.06mr = \frac{12}{2} = 6 \, \text{cm} = 0.06 \, \text{m} The area AA of the circular patch is: A=πr2=π(0.06)2=0.0113m2A = \pi r^2 = \pi (0.06)^2 = 0.0113 \, \text{m}^2

Step 5: Average Power Calculation

The average power is the product of the time-averaged Poynting vector and the area of the patch: Pavg=SAP_{\text{avg}} = \langle S \rangle A

Substituting for S\langle S \rangle: Pavg=12η0E02AP_{\text{avg}} = \frac{1}{2\eta_0} |E_0|^2 \cdot A Now plug in the known values: Pavg=12×377(50×103)20.0113P_{\text{avg}} = \frac{1}{2 \times 377} (50 \times 10^3)^2 \cdot 0.0113

Step 6: Compute the Power

Let's compute this step.The average power crossing the circular patch is approximately 37.5kW37.5 \, \text{kW}.

Would you like more details on any specific step? Here are five related questions to expand on this topic:

  1. How is the Poynting vector related to the power carried by an electromagnetic wave?
  2. What is the significance of the intrinsic impedance in wave propagation?
  3. How would the calculation change if the medium was not free space (non-vacuum)?
  4. What would be the effect of changing the frequency of the wave on the average power?
  5. How does the diameter of the circular patch affect the average power calculation?

Tip: In problems involving electromagnetic waves, always pay attention to the medium's properties, as they affect the wave's behavior significantly.

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Math Problem Analysis

Mathematical Concepts

Electromagnetism
Wave Propagation
Poynting Vector
Electric Field Intensity

Formulas

E = 50sin(10^8 t - 2z) ak (kV/m)
S = (1/μ0)(E × H)
⟨S⟩ = (1/2η)|E0|^2
A = πr^2

Theorems

Poynting Theorem
Intrinsic Impedance of Free Space

Suitable Grade Level

Undergraduate - Electrical Engineering