JEE MAIN - Physics (2023 - 8th April Morning Shift - No. 5)
Dimension of $$\frac{1}{\mu_{0} \in_{0}}$$ should be equal to
$$\mathrm{T}^{2} / \mathrm{L}^{2}$$
$$\mathrm{T} / \mathrm{L}$$
$$\mathrm{L}^{2} / \mathrm{T}^{2}$$
$$\mathrm{L} / \mathrm{T}$$
Explanation
The term $\frac{1}{\mu_{0} \epsilon_{0}}$ appears in the formula for the speed of light $c$, which is:
$c = \sqrt{\frac{1}{\mu_{0} \epsilon_{0}}}$
where $\mu_{0}$ is the permeability of free space and $\epsilon_{0}$ is the permittivity of free space.
The speed of light $c$ has dimensions of length over time ($\mathrm{L} / \mathrm{T}$). Therefore, the term $\frac{1}{\mu_{0} \epsilon_{0}}$ has dimensions equal to the square of the speed of light, which is $\mathrm{L}^{2} / \mathrm{T}^{2}$.
$c = \sqrt{\frac{1}{\mu_{0} \epsilon_{0}}}$
where $\mu_{0}$ is the permeability of free space and $\epsilon_{0}$ is the permittivity of free space.
The speed of light $c$ has dimensions of length over time ($\mathrm{L} / \mathrm{T}$). Therefore, the term $\frac{1}{\mu_{0} \epsilon_{0}}$ has dimensions equal to the square of the speed of light, which is $\mathrm{L}^{2} / \mathrm{T}^{2}$.
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