The speed of electromagnetic waves in a medium is related to its relative permeability (\( \mu_r \)) and relative permittivity (\( \varepsilon_r \)) by the equation:
\[ \varepsilon_r \mu_r = \frac{c^2}{v^2}, \]
where:
- \( c = 3 \times 10^8 \, \text{ms}^{-1} \) (speed of light in vacuum),
- \( v = 1.5 \times 10^8 \, \text{ms}^{-1} \) (speed of light in the medium),
- \( \mu_r = 2.0 \) (relative permeability of the medium).
Substituting the given values:
\[ \varepsilon_r \times 2 = \frac{(3 \times 10^8)^2}{(1.5 \times 10^8)^2}. \]
Simplify:
\[ \varepsilon_r \times 2 = \frac{9 \times 10^{16}}{2.25 \times 10^{16}}. \]
\[ \varepsilon_r \times 2 = 4. \]
\[ \varepsilon_r = 2. \]
Final Answer: \( \varepsilon_r = 2 \) (Option 4)
To find the relative permittivity of the medium, we begin with the relationship between the speed of electromagnetic waves, permittivity, and permeability. The speed of electromagnetic waves in a medium is given by:
\(v = \frac{c}{\sqrt{\mu_r \cdot \epsilon_r}}\)
where:
We need to find \(\epsilon_r\). Rearrange the formula to solve for \(\epsilon_r\):
\(\epsilon_r = \frac{c^2}{v^2 \cdot \mu_r}\)
Substitute the known values:
\(\epsilon_r = \frac{(3 \times 10^8)^2}{(1.5 \times 10^8)^2 \cdot 2}\)
Calculate the squared terms:
Substitute back into the formula:
\(\epsilon_r = \frac{9 \times 10^{16}}{2.25 \times 10^{16} \cdot 2}\)
Simplify the expression:
\(\epsilon_r = \frac{9}{4.5} = 2\)
Thus, the relative permittivity of the medium is \(2\).
The correct answer is: 2
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