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.

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:
\[
E = 50 \sin(10^8 t - 2z) \, \hat{a}_k \, \text{kV/m}
\]
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** \(\langle \mathbf{S} \rangle\), which represents power per unit area. The Poynting vector is expressed as:
\[
\mathbf{S} = \frac{1}{\mu_0} (\mathbf{E} \times \mathbf{H})
\]
However, for the average power, we use:
\[
\langle S \rangle = \frac{1}{2\eta} |E_0|^2
\]
where:
- \(E_0\) is the amplitude of the electric field.
- \(\eta\) is the intrinsic impedance of the medium.
- \(\mu_0 = 4\pi \times 10^{-7} \, \text{H/m}\) is the permeability of free space.

For a **non-magnetic medium**, the intrinsic impedance \(\eta\) is approximately the intrinsic impedance of free space:
\[
\eta_0 = 377 \, \Omega
\]

### Step 3: Electric Field Amplitude
The amplitude of the electric field is given as \(E_0 = 50 \, \text{kV/m} = 50 \times 10^3 \, \text{V/m}\).

### Step 4: Area of the Circular Patch
The patch is circular with a diameter of 12 cm, so the radius \(r\) is:
\[
r = \frac{12}{2} = 6 \, \text{cm} = 0.06 \, \text{m}
\]
The area \(A\) of the circular patch is:
\[
A = \pi r^2 = \pi (0.06)^2 = 0.0113 \, \text{m}^2
\]

### Step 5: Average Power Calculation
The average power is the product of the time-averaged Poynting vector and the area of the patch:
\[
P_{\text{avg}} = \langle S \rangle A
\]

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

### Step 6: Compute the Power
Let's compute this step.The average power crossing the circular patch is approximately \( 37.5 \, \text{kW} \).

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.

Learn how to calculate the average power crossing a circular patch in a non-magnetic medium for a plane wave with electric field intensity E=50sin(108t-2z)ak kV/m. This problem covers key concepts like the Poynting vector, electric field intensity, and the intrinsic impedance of free space. Suitable for undergraduate students in electrical engineering.

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