Question:
$\mu_0$ and $\varepsilon_0$ denote the permeability and permittivity of free space, the dimensions of $\mu_0 \varepsilon_0$ are
$\mu_0$ and $\varepsilon_0$ denote the permeability and permittivity of free space, the dimensions of $\mu_0 \varepsilon_0$ are
Updated On: Apr 2, 2024
- ${[LT^{-1}]}$
- ${[L^{-2}T^{2}]}$
- ${[M^{-1}L^{-3} T^{2}]}$
- ${[M^{-1}L^{-3} T^{2}A^2]}$
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The Correct Option is B
Solution and Explanation
Speed of light in vacuum is given by
$c = \frac{1}{\sqrt{\mu_0 \varepsilon_0}}$ or $ \mu_0 \varepsilon_0 = \frac{1}{c^2}$
${ \therefore \:\:\:\: [\mu_0 \varepsilon_0 ]= \frac{1}{[LT^{-1}]^2 } = [L^{-2} T^2]}$
$c = \frac{1}{\sqrt{\mu_0 \varepsilon_0}}$ or $ \mu_0 \varepsilon_0 = \frac{1}{c^2}$
${ \therefore \:\:\:\: [\mu_0 \varepsilon_0 ]= \frac{1}{[LT^{-1}]^2 } = [L^{-2} T^2]}$
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