Q 11-01-169JEE MainJEE Main 2022 (28 Jul, Shift 1)Easy
The dimensions of $\dfrac{B^2}{\mu_0}$ will be (if $\mu_0$: permeability of free space and $B$ : magnetic field)
Answer: (C) $[\text{ML}^{-1}\text{T}^{-2}]$
$\dfrac{B^2}{2\mu_0}$ is the energy density of a magnetic field, so $\dfrac{B^2}{\mu_0}$ has the dimensions of energy per volume:
$$\frac{[\text{ML}^2\text{T}^{-2}]}{[\text{L}^3]} = [\text{ML}^{-1}\text{T}^{-2}]$$
Solution by Sreeraj P, M.Sc Physics