The dimensions of $$\frac{B^2}{\mu_0}$$ will be (if $$\mu_0$$: permeability of free space and $$B$$: magnetic field)

We need to find the dimensions of $$\frac{B^2}{\mu_0}$$.

From the Lorentz force equation $$F = qvB$$, we get:

$$[B] = \frac{[F]}{[q][v]} = \frac{MLT^{-2}}{AT \cdot LT^{-1}} = MT^{-2}A^{-1}$$

Meanwhile, using $$B = \mu_0 \frac{I}{2\pi r}$$ leads to

$$[\mu_0] = \frac{[B][L]}{[I]} = \frac{MT^{-2}A^{-1} \cdot L}{A} = MLT^{-2}A^{-2}$$

Combining these results gives the dimension of $$\frac{B^2}{\mu_0}$$ as

$$\left[\frac{B^2}{\mu_0}\right] = \frac{[MT^{-2}A^{-1}]^2}{MLT^{-2}A^{-2}} = \frac{M^2T^{-4}A^{-2}}{MLT^{-2}A^{-2}} = ML^{-1}T^{-2}$$

This matches the dimension of pressure or energy density, consistent with the interpretation of $$\frac{B^2}{2\mu_0}$$ as magnetic energy density. Therefore, the correct answer is Option C: $$[ML^{-1}T^{-2}]$$.