JEE MAIN - Physics (2022 - 25th June Evening Shift - No. 13)
The electromagnetic waves travel in a medium at a speed of 2.0 $$\times$$ 108 m/s. The relative permeability of the medium is 1.0. The relative permittivity of the medium will be :
2.25
4.25
6.25
8.25
Explanation
The speed of electromagnetic waves in a medium is given by the formula:
$$v = \frac{1}{\sqrt{\mu \varepsilon}}$$
where $\mu$ and $\varepsilon$ are the absolute permeability and absolute permittivity of the medium, respectively.
Given that $\mu = \mu_0 \mu_r$ and $\varepsilon = \varepsilon_0 \varepsilon_r$, we can rewrite the above equation as :
$$v = \frac{1}{\sqrt{\mu_0 \mu_r \varepsilon_0 \varepsilon_r}}$$
which simplifies to :
$$v = \frac{1}{\sqrt{\mu_0 \varepsilon_0}} \times \frac{1}{\sqrt{\mu_r \varepsilon_r}}$$
Substituting $c$ for $\frac{1}{\sqrt{\mu_0 \varepsilon_0}}$ (the speed of light in a vacuum), we get:
$$v = \frac{c}{\sqrt{\mu_r \varepsilon_r}}$$
We can rearrange this equation to solve for $\varepsilon_r$ :
$$\varepsilon_r = \frac{c^2}{v^2 \mu_r} = \frac{(3 \times 10^8 m/s)^2}{(2 \times 10^8 m/s)^2 \times 1} = 2.25$$
So, the relative permittivity of the medium is 2.25.
$$v = \frac{1}{\sqrt{\mu \varepsilon}}$$
where $\mu$ and $\varepsilon$ are the absolute permeability and absolute permittivity of the medium, respectively.
Given that $\mu = \mu_0 \mu_r$ and $\varepsilon = \varepsilon_0 \varepsilon_r$, we can rewrite the above equation as :
$$v = \frac{1}{\sqrt{\mu_0 \mu_r \varepsilon_0 \varepsilon_r}}$$
which simplifies to :
$$v = \frac{1}{\sqrt{\mu_0 \varepsilon_0}} \times \frac{1}{\sqrt{\mu_r \varepsilon_r}}$$
Substituting $c$ for $\frac{1}{\sqrt{\mu_0 \varepsilon_0}}$ (the speed of light in a vacuum), we get:
$$v = \frac{c}{\sqrt{\mu_r \varepsilon_r}}$$
We can rearrange this equation to solve for $\varepsilon_r$ :
$$\varepsilon_r = \frac{c^2}{v^2 \mu_r} = \frac{(3 \times 10^8 m/s)^2}{(2 \times 10^8 m/s)^2 \times 1} = 2.25$$
So, the relative permittivity of the medium is 2.25.
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