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 :

• A. \( L T^2 \)
• B. \( L^2 T^{-2} \)
• C. \( T^2 / L \)
• D. \( T^2 / L^2 \)

Hint:

The dimensions of permeability and permittivity help determine the speed of light in a vacuum, and their relationship is key in electromagnetic theory.

Answer 1

The Correct Option is B

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., \( c^2 \)).

Answer 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 \( L^2 T^2 \).