Question

If the dimensional unit of magnetic permeability ($ \mu $) is given by $ [MLT^{-2}I^{-2}] $, then the dimensional unit of electric permittivity ($ \epsilon $) will be

• A. \( [ML^3T^{-4}I^{-2}] \)
• B. \( [M^{-1}L^{-3}T^4I^2] \)
• C. \( [M^{-1}L^3T^{-4}I^{-2}] \)
• D. \( [ML^3T^{-4}I^{-2}] \)

Hint:

When solving for dimensional formulas, use the relationships between physical quantities like the speed of light, magnetic permeability, and electric permittivity.

Answer 1

The Correct Option is C

Step 1: Understanding the relationship between electric permittivity and magnetic permeability.
The electric permittivity \( \epsilon \) and magnetic permeability \( \mu \) are related through the speed of light \( c \) in a vacuum: \[ c^2 = \frac{1}{\mu \epsilon} \] The speed of light has the dimensional formula \( [LT^{-1}] \).
Step 2: Dimensional analysis.
We are given that the dimensional unit of magnetic permeability \( \mu \) is \( [MLT^{-2}I^{-2}] \). Now, substituting into the equation \( c^2 = \frac{1}{\mu \epsilon} \), we can solve for the dimensional formula of \( \epsilon \). 
Using dimensional analysis, we get the formula for electric permittivity as \( [M^{-1}L^3T^{-4}I^{-2}] \). Thus, the correct answer is 
(C) \( [M^{-1}L^3T^{-4}I^{-2}] \).