Question

The dimensional formula of magnetic permeability \(\mu_0\) is:

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

Hint:

To find the dimensional formula of magnetic permeability \(\mu_0\), use the force between current-carrying wires or the relationship \( B = \mu_0 H \). The dimensions often involve \( A^{-2} \) due to the inverse dependence on current squared.

Answer 1

The Correct Option is A

Step 1: Determine the quantity.
Given the question's incompleteness and the presence of current \( A \) with negative exponents in the options, it is inferred that the dimensional formula for magnetic permeability \(\mu_0\), a fundamental constant in electromagnetism, is required.Step 2: Calculate the dimensions of \(\mu_0\).
The formula for the force per unit length between two parallel current-carrying wires is given by:\[\frac{F}{\ell} = \frac{\mu_0 I_1 I_2}{2\pi d}\]Rearranging for \(\mu_0\):\[\mu_0 = \frac{F}{\ell} \cdot \frac{2\pi d}{I_1 I_2}\]The dimensions of the components are:- Force per unit length: \([F/\ell] = [M T^{-2}]\)
- Distance \( d \): \([d] = [L]\)
- Current \( I \): \([I] = [A]\)
Therefore, the dimensional formula for \(\mu_0\) is:\[[\mu_0] = \frac{[M T^{-2}] [L]}{[A]^2} = [M L T^{-2} A^{-2}]\]Step 3: Correlate with the provided options.
The derived dimensional formula, \([M L T^{-2} A^{-2}]\), corresponds to option (A).