Question

Electromagnetic waves travel in a medium with speed \( 1.5 \times 10^8 \) m/s. The relative permeability of the medium is \( 2.0 \). The relative permittivity will be:

• A. 5
• B. 1
• C. 4
• D. 2

Hint:

The speed of light in any medium is given by \( v = \frac{c}{\sqrt{\mu_r \varepsilon_r}} \), where \( \mu_r \) and \( \varepsilon_r \) are the relative permeability and permittivity.

Answer 1

The Correct Option is D

Step 1: {Using the Speed of Light Formula}
\[ v = \frac{c}{\sqrt{\mu_r \varepsilon_r}} \] Step 2: {Solving for \( \varepsilon_r \)}
\[ 1.5 \times 10^8 = \frac{3 \times 10^8}{\sqrt{2 \varepsilon_r}} \] \[ \sqrt{2 \varepsilon_r} = 2 \] \[ \varepsilon_r = 2 \] Thus, the correct answer is \( 2 \).

Answer 2

Step 1: Use the formula for speed of electromagnetic waves in a medium
The speed of electromagnetic waves in a medium is given by:
\( v = \frac{1}{\sqrt{\mu \varepsilon}} \)
Where:
- \( \mu = \mu_0 \mu_r \) is the absolute permeability
- \( \varepsilon = \varepsilon_0 \varepsilon_r \) is the absolute permittivity
- \( \mu_r \) = relative permeability
- \( \varepsilon_r \) = relative permittivity
- \( c = \frac{1}{\sqrt{\mu_0 \varepsilon_0}} = 3 \times 10^8 \) m/s (speed of light in vacuum)

So we can write:
\( v = \frac{c}{\sqrt{\mu_r \varepsilon_r}} \)

Step 2: Plug in known values
Given:
- \( v = 1.5 \times 10^8 \) m/s
- \( c = 3 \times 10^8 \) m/s
- \( \mu_r = 2.0 \)

Now use the equation:
\( \frac{v}{c} = \frac{1}{\sqrt{\mu_r \varepsilon_r}} \)
\( \frac{1.5 \times 10^8}{3 \times 10^8} = \frac{1}{\sqrt{2 \cdot \varepsilon_r}} \)
\( \frac{1}{2} = \frac{1}{\sqrt{2 \cdot \varepsilon_r}} \)
Cross-multiplying:
\( \sqrt{2 \cdot \varepsilon_r} = 2 \)
Squaring both sides:
\( 2 \cdot \varepsilon_r = 4 \)
\( \varepsilon_r = \frac{4}{2} = 2 \)

Step 3: Final Answer
The relative permittivity of the medium is:
2