What is the dimensional formula of \( \frac{1}{\mu_0 \epsilon_0} \) (where \( \mu_0 \) is permeability and \( \epsilon_0 \) is permittivity of free space)?
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To calculate dimensional formulas involving constants, break down each constant's dimensional formula and perform the necessary arithmetic operations.
The expression \( \frac{1}{\mu_0 \epsilon_0} \) is defined using the permeability of free space \( \mu_0 \) and the permittivity of free space \( \epsilon_0 \).- The dimensional formula for \( \mu_0 \) is \( \text{M}^{-1} \text{L}^{-3} \text{T}^4 \text{A}^2 \), with \( A \) denoting electric current.- The dimensional formula for \( \epsilon_0 \) is \( \text{M}^{-1} \text{L}^{-3} \text{T}^4 \text{A}^2 \).Consequently, the dimensional formula for \( \frac{1}{\mu_0 \epsilon_0} \) is derived as follows:\[\left( \frac{1}{\mu_0 \epsilon_0} \right) = \left( \text{M}^{-1} \text{L}^{-3} \text{T}^4 \text{A}^2 \right)^{-1} = \text{M} \text{L}^3 \text{T}^{-4} \text{A}^{-2}\]This simplifies to the dimensional formula of \( \frac{1}{\mu_0 \epsilon_0} \) as \( \text{ML}^2 \text{T}^{-2} \). Therefore, option (4) is the correct answer.
