JEE MAIN - Physics (2022 - 28th July Morning Shift - No. 1)
The dimensions of $$\left(\frac{\mathrm{B}^{2}}{\mu_{0}}\right)$$ will be :
(if $$\mu_{0}$$ : permeability of free space and $$B$$ : magnetic field)
$$\left[\mathrm{M}\, \mathrm{L}^2 \,\mathrm{T}^{-2}\right]$$
$$\left[\mathrm{M} \,\mathrm{L} \,\mathrm{T}^{-2}\right]$$
$$\left[\mathrm{M} \,\mathrm{L}^{-1} \,\mathrm{~T}^{-2}\right]$$
$$\left[\mathrm{M} \,\mathrm{L}^{2} \mathrm{~T}^{-2} \mathrm{~A}^{-1}\right]$$
설명
$$\left[ {{{{B^2}} \over {{\mu _0}}}} \right]=$$ [Energy density]
$$ = {{M{L^2}{T^{ - 2}}} \over {{L^3}}} = M{L^{ - 1}}{T^{ - 2}}$$