Q 11-01-202JEE MainJEE Main 2020 (7 Jan, Shift 2)Easy
Given, $B$ is magnetic field induction and $\mu_0$ is the magnetic permeability of vacuum. The dimension of $\dfrac{B^2}{2\mu_0}$ is:
Answer: (D) $ML^{-1}T^{-2}$
$\dfrac{B^2}{2\mu_0}$ is the energy density (energy per unit volume) of a magnetic field:
$$\left[\frac{B^2}{2\mu_0}\right] = \frac{ML^2T^{-2}}{L^3} = ML^{-1}T^{-2}$$
Solution by Sreeraj P, M.Sc Physics