The electromagnetic waves… - JEE Main 2022 Physics PYQ

JEE Main 2022 — Physics Electromagnetism

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The electromagnetic waves travel in a medium at a speed of $2.0\times {10}^{8}m{s}^{-1}$ . The relative permeability of the medium is $1.0$ . The relative permittivity of the medium will be

Official previous-year question

Held on 25 Jun 2022 · Verified 6 Jul 2026.

Options

A
$2.25$
B
$4.25$
C
$6.25$
D
$8.25$

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Solution

Here, $v=2\times {10}^{8} \text{m} {\text{s}}^{-1}$ and ${\mu }_{r}=1$

The speed of electromagnetic waves in a medium is given by $v=\frac{1}{\sqrt{\mu \epsilon }}...(i)$

where $\mu$ and $\epsilon$ are absolute permeability and absolute permittivity of the medium.

Now, $\mu ={\mu }_{0}{\mu }_{r}$ and $\epsilon ={\epsilon }_{0}{\epsilon }_{r}$

Then, we have,

$v=\frac{1}{\sqrt{{\mu }_{0}{\mu }_{r}\times {\epsilon }_{0}{\epsilon }_{r}}}=\frac{1}{\sqrt{{\mu }_{0}{\epsilon }_{0}}}\times \frac{1}{\sqrt{{\mu }_{r}{\epsilon }_{r}}}=\frac{c}{\sqrt{{\mu }_{r}{\epsilon }_{r}}}$

$or {\epsilon }_{r}=\frac{{c}^{2}}{{v}^{2}{\mu }_{r}}=\frac{{(3\times {10}^{8})}^{2}}{{(2\times {10}^{8})}^{2}}=\frac{9}{4}=2.25$

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Solution

Here, $v=2\times {10}^{8} \text{m} {\text{s}}^{-1}$ and ${\mu }_{r}=1$

The speed of electromagnetic waves in a medium is given by $v=\frac{1}{\sqrt{\mu \epsilon }}...(i)$

where $\mu$ and $\epsilon$ are absolute permeability and absolute permittivity of the medium.

Now, $\mu ={\mu }_{0}{\mu }_{r}$ and $\epsilon ={\epsilon }_{0}{\epsilon }_{r}$

Then, we have,

$or {\epsilon }_{r}=\frac{{c}^{2}}{{v}^{2}{\mu }_{r}}=\frac{{(3\times {10}^{8})}^{2}}{{(2\times {10}^{8})}^{2}}=\frac{9}{4}=2.25$

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Solution

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Options

Did you get this right?

Solution

Your note

JEE Main Electromagnetism in other years

More JEE Main Electromagnetism questions

Other Physics topics for JEE Main

Mechanics

Modern Physics

Optics

Thermodynamics

Waves & Oscillations

Keep practicing Electromagnetism