A wave is propagating in a medium of electric dielectric constant `2` and relative magnetic permeability `50`. The wave impeldance of such a medium is

Options

A. `5 Omega`
B. `376.6 Omega`
C. `1883 Omega`
D. `3776 Omega`

Explanation

To find the wave impedance of a medium with a given dielectric constant and relative magnetic permeability, we can use the formula for wave impedance \( Z \):

\[ Z = \sqrt{\frac{\mu_r}{\epsilon_r} \cdot \frac{\mu_0}{\epsilon_0}} \]

Where:
- \( \mu_r \) is the relative magnetic permeability,
- \( \epsilon_r \) is the relative permittivity (dielectric constant),
- \( \mu_0 \) is the permeability of free space (approximately \( 4\pi \times 10^{-7} \, \text{H/m} \)),
- \( \epsilon_0 \) is the permittivity of free space (approximately \( 8.854 \times 10^{-12} \, \text{F/m} \)).

Given:
- \( \epsilon_r = 2 \)
- \( \mu_r = 50 \)

### Step 1: Substitute the values into the wave impedance formula

\[ Z = \sqrt{\frac{50}{2} \cdot \frac{\mu_0}{\epsilon_0}} \]

### Step 2: Simplify the fraction inside the square root

\[ Z = \sqrt{25 \cdot \frac{\mu_0}{\epsilon_0}} \]

### Step 3: Calculate \( \frac{\mu_0}{\epsilon_0} \)

The value of \( \frac{\mu_0}{\epsilon_0} \) is known to be approximately \( 376.6 \, \Omega \).

### Step 4: Substitute this value back into the equation

\[ Z = \sqrt{25 \cdot 376.6} \]

### Step 5: Calculate the product

\[ Z = \sqrt{9415} \]

### Step 6: Calculate the square root

\[ Z \approx 97 \, \Omega \]

### Final Answer

The wave impedance of the medium is approximately \( 97 \, \Omega \).

---