An electromagnetic wave of frequency 5 GHz, is travelling in a medium whose relative electric permittivity and relative magnetic permeability both are 2. Its velocity in this medium is _______` xx 10^7 m//s`

Answer

To find the velocity of an electromagnetic wave traveling in a medium with given properties, we can follow these steps:

### Step 1: Understand the given data
We have the frequency of the electromagnetic wave \( f = 5 \text{ GHz} \) and both the relative electric permittivity \( \epsilon_r = 2 \) and relative magnetic permeability \( \mu_r = 2 \).

### Step 2: Convert frequency to Hz
Convert the frequency from GHz to Hz:
\[ f = 5 \text{ GHz} = 5 \times 10^9 \text{ Hz} \]

### Step 3: Use the formula for velocity in a medium
The velocity \( V \) of an electromagnetic wave in a medium is given by:
\[ V = \frac{1}{\sqrt{\mu \epsilon}} \]
Where:
- \( \mu = \mu_0 \mu_r \)
- \( \epsilon = \epsilon_0 \epsilon_r \)

### Step 4: Substitute the values of \( \mu \) and \( \epsilon \)
Substituting the expressions for \( \mu \) and \( \epsilon \):
\[ V = \frac{1}{\sqrt{\mu_0 \mu_r \cdot \epsilon_0 \epsilon_r}} \]

### Step 5: Factor out constants
We can express this in terms of the speed of light \( c \):
\[ V = \frac{1}{\sqrt{\mu_r \epsilon_r}} \cdot \frac{1}{\sqrt{\mu_0 \epsilon_0}} = \frac{c}{\sqrt{\mu_r \epsilon_r}} \]

### Step 6: Substitute the values of \( \mu_r \) and \( \epsilon_r \)
Now substituting \( \mu_r = 2 \) and \( \epsilon_r = 2 \):
\[ V = \frac{c}{\sqrt{2 \cdot 2}} = \frac{c}{\sqrt{4}} = \frac{c}{2} \]

### Step 7: Use the value of \( c \)
The speed of light \( c \) is approximately \( 3 \times 10^8 \text{ m/s} \):
\[ V = \frac{3 \times 10^8}{2} = 1.5 \times 10^8 \text{ m/s} \]

### Step 8: Express the final answer in the required format
To express this in the format \( xx \times 10^7 \text{ m/s} \):
\[ V = 15 \times 10^7 \text{ m/s} \]

### Final Answer
The velocity of the electromagnetic wave in the medium is \( 15 \times 10^7 \text{ m/s} \).

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