Question

Electromagnetic waves travel in a medium with a speed of $1.5 \times 10^8 \, \text{ms}^{-1}$. The relative permeability of the medium is $2.0$. The relative permittivity will be:

• A. 5
• B. 4
• C. 1
• D. 2

Answer 1

Answer: (D)

2

Answer 2

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)