Question:
If $ \mu_0 $ and $ \epsilon_0 $ are the permeability and permittivity of free space, respectively, then the dimension of $ \left( \frac{1}{\mu_0 \epsilon_0} \right) $ is :
If $ \mu_0 $ and $ \epsilon_0 $ are the permeability and permittivity of free space, respectively, then the dimension of $ \left( \frac{1}{\mu_0 \epsilon_0} \right) $ is :
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The dimensions of permeability and permittivity help determine the speed of light in a vacuum, and their relationship is key in electromagnetic theory.
Updated On: Jan 14, 2026
- \( L T^2 \)
\( L^2 T^{-2} \)
- \( T^2 / L \)
- \( T^2 / L^2 \)
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The Correct Option is B
Approach Solution - 1
We are asked to find the dimension of \( \left(\dfrac{1}{\mu_0 \epsilon_0}\right) \), where \( \mu_0 \) and \( \epsilon_0 \) are the permeability and permittivity of free space, respectively.
Concept Used:
We use the relation between the speed of light \( c \), the permeability \( \mu_0 \), and the permittivity \( \epsilon_0 \):
\[ c = \frac{1}{\sqrt{\mu_0 \epsilon_0}} \]Hence,
\[ \frac{1}{\mu_0 \epsilon_0} = c^2 \]The dimension of speed \( c \) is \( [L\,T^{-1}] \), so the dimension of \( c^2 \) is \( [L^2\,T^{-2}] \).
Step-by-Step Solution:
Step 1: Write the dimensional formula of \( \mu_0 \) and \( \epsilon_0 \):
\[ [\mu_0] = [M^1 L^1 T^{-2} A^{-2}] \] \[ [\epsilon_0] = [M^{-1} L^{-3} T^{4} A^{2}] \]Step 2: Multiply \( \mu_0 \epsilon_0 \):
\[ [\mu_0 \epsilon_0] = [M^{1-1} L^{1-3} T^{-2+4} A^{-2+2}] = [L^{-2} T^{2}] \]Step 3: Take the reciprocal:
\[ \left[\frac{1}{\mu_0 \epsilon_0}\right] = [L^2 T^{-2}] \]Final Computation & Result:
The dimension of \( \left(\dfrac{1}{\mu_0 \epsilon_0}\right) \) is:
\[ \boxed{[L^2 T^{-2}]} \]It represents the square of velocity (i.e., \( c^2 \)).
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Approach Solution -2
Using the formula,
\[
C = \frac{1}{\sqrt{\mu_0 \epsilon_0}} \Rightarrow 1 = C^2 = L T^{-2}
\]
Thus, the dimension of \( \frac{1}{\mu_0 \epsilon_0} \) is \( L^2 T^2 \).
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