Electromagnetic waves travel in a medium with speed of 1.5 × 10^8 ~m ~s^ -1 . The relative permeability of the medium is 2.0 . The relative permittivity will be:

Correct answer: A. 2 aligned & _ m × μ_ m _0 × μ_0 = 1 v^2 1 c^2 & _ r × μ_ r = c^2 v^2 & _ r × 2=((3 × 10^8)^2)/((1.5 × 10^8)^2) & _ r × 2=4 & _ r =2 aligned

JEE Main 20246 AprilMorning Shift

Exam: JEE Main
Year: 2024
Date: 6 April
Shift: Morning Shift
Subject: Physics
Chapter: Electromagnetic Waves

JEE Main 2024 · 6 April · Morning Shift · Physics · Electromagnetic Waves

Electromagnetic waves travel in a medium with speed of 1.5 × 10^8...

Electromagnetic waves travel in a medium with speed of

1.5 \times 10^8 \mathrm{~m} \mathrm{~s}^{-1}

. The relative permeability of the medium is 2.0 . The relative permittivity will be:

Options

Correct answer

A. 2

Step-by-step solution

\begin{aligned} & \frac{\varepsilon_{\mathrm{m}} \times \mu_{\mathrm{m}}}{\varepsilon_0 \times \mu_0}=\frac{\frac{1}{\mathrm{v}^2}}{\frac{1}{\mathrm{c}^2}} \\ & \varepsilon_{\mathrm{r}} \times \mu_{\mathrm{r}}=\frac{\mathrm{c}^2}{\mathrm{v}^2} \\ & \varepsilon_{\mathrm{r}} \times 2=\frac{\left(3 \times 10^8\right)^2}{\left(1.5 \times 10^8\right)^2} \\ & \varepsilon_{\mathrm{r}} \times 2=4 \\ & \varepsilon_{\mathrm{r}}=2\end{aligned}

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