The dimensions of $\left(\frac{\mathrm{B}^{2}}{\mu_{0}}\right)$ will be : (if $\mu_{0}$ : permeability of free space and B : magnetic field)

(C) $\left[\mathrm{M} \mathrm{L}^{-1} \mathrm{~T}^{-2}\right]$

(A) $\left[\mathrm{M} \mathrm{L}^{2} \mathrm{~T}^{-2}\right]$

(B) $\left[\mathrm{M} \mathrm{L} \mathrm{T}^{-2}\right]$

(D) $\left[\mathrm{M} \mathrm{L}^{2} \mathrm{~T}^{-2} \mathrm{~A}^{-1}\right]$

JEE Main · 2022 · 28 Jul 2022 · Shift 1 · Units & Measurement

The dimensions of

(\frac{B ^{2}}{μ _{0}})

will be :
(if

μ_{0}

: permeability of free space and
B : magnetic field)

Answer: (C)

The quantity

\frac{B ^{2}}{μ _{0}}

represents energy density (energy per unit volume). The magnetic energy density is

u = \frac{B ^{2}}{2 μ _{0}}

, so

\frac{B ^{2}}{μ _{0}}

has the same dimensions as energy density. Energy has dimensions

[M L^{2} T^{- 2}]

, so energy per unit volume has dimensions

\frac{[ M L ^{2} T ^{- 2} ]}{[ L ^{3} ]} = [M L^{- 1} T^{- 2}]

JEE Main · 2022 · 28 Jul 2022 · Shift 1 · Single correct option