Question:
The velocity of electromagnetic waves in vacuum is given by
The velocity of electromagnetic waves in vacuum is given by
Show Hint
Remember that the velocity of light in a vacuum can be calculated using the constants $\mu_0$ and $\epsilon_0$, which are fundamental to electromagnetism.
Updated On: May 31, 2026
- $1 / \sqrt{\mu_0 \epsilon_0}$
- $\sqrt{\mu_0 \epsilon_0}$
- $\mu_0 \epsilon_0$
- $1 / (\mu_0 \epsilon_0)$
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The Correct Option is A
Solution and Explanation
Step 1: Concept
The velocity of electromagnetic waves in vacuum is a fundamental concept derived from Maxwell's equations. It relates to the permeability ($\mu_0$) and permittivity ($\epsilon_0$) of free space.
Step 2: Meaning
$\mu_0$ represents the magnetic permeability of free space, while $\epsilon_0$ denotes the electric permittivity of free space. These constants are intrinsic properties of the vacuum and are used to describe how electromagnetic fields behave in a vacuum.
Step 3: Analysis
The velocity ($v$) of an electromagnetic wave in a medium is given by the equation: \[v = \frac{1}{\sqrt{\mu_0 \epsilon_0}}\] This formula comes from the relationship between the speed of light and the properties of the medium. In vacuum, this simplifies to the velocity of electromagnetic waves. Option A) $1 / \sqrt{\mu_0 \epsilon_0}$ matches exactly with the derived equation for the velocity of electromagnetic waves in a vacuum. Options B), C), and D) do not match the correct form: Option B) $\sqrt{\mu_0 \epsilon_0}$ would give an incorrect velocity. Option C) $\mu_0 \epsilon_0$ is the product of permeability and permittivity, which does not represent a velocity. Option D) $1 / (\mu_0 \epsilon_0)$ would also yield an incorrect result.
Step 4: Conclusion
The correct expression for the velocity of electromagnetic waves in vacuum is derived from the inverse square root of the product of $\mu_0$ and $\epsilon_0$. Final Answer: (A)
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