Question

The dimensions of \( (\mu \epsilon)^{-1} \), where \( \epsilon \) is permittivity and \( \mu \) is permeability of a medium, are:

• A. \( [M^0 L^1 T^{-1}] \)
• B. \( [M^0 L^2 T^{-2}] \)
• C. \( [M^1 L^2 T^{-2}] \)
• D. \( [M^1 L^{-1} T^{1}] \)

Hint:

To calculate the dimensions of a product or quotient of physical quantities, multiply or divide their respective dimensions. For inverse quantities, simply invert the dimensions.

Answer 1

The Correct Option is B

Solution: Determine Dimensions of \( (\mu \epsilon)^{-1} \)

To find the dimensions of \( (\mu \epsilon)^{-1} \), we first establish the dimensions of permittivity (\( \epsilon \)) and permeability (\( \mu \)).

Step 1: Obtain formulas for permittivity and permeability:

\[ \epsilon = \frac{1}{\mu_0 c^2}, \quad \mu = \frac{1}{\epsilon_0 c^2} \]

Step 2: Deduce dimensions of permittivity:

\[ [\epsilon] = [M^{-1} L^{-3} T^4 A^2] \]

Step 3: Deduce dimensions of permeability:

\[ [\mu] = [M L T^{-2} A^{-2}] \]

Step 4: Compute dimensions of the product \( \mu \epsilon \):

\[ [\mu \epsilon] = [M L T^{-2} A^{-2}] \times [M^{-1} L^{-3} T^4 A^2] = [M^0 L^{-2} T^2 A^0] \]

Step 5: Determine dimensions of \( (\mu \epsilon)^{-1} \):

\[ (\mu \epsilon)^{-1} = [M^0 L^2 T^{-2} A^0] = [L^2 T^{-2}] \]

Conclusion: The dimensions of \( (\mu \epsilon)^{-1} \) are:

\[ [L^2 T^{-2}] \]