The dimensions of B 2 μ 0 will be (if μ 0 : permeability of free space and B   : magnetic…

The dimensions of $(\frac{B^{2}}{μ_{0}})$ will be

(if $μ_{0}$ : permeability of free space and $B :$ magnetic field)

$[{ML}^{2} T^{- 2}]$
$[{MLT}^{- 2}]$
$[{ML}^{- 1} T^{- 2}]$
$[{MLL}^{- 2} A^{- 1}]$

Solution

Energy density in magnetic field is given by $u = \frac{B^{2}}{2 μ_{0}}$ . Therefore, dimensions of $\frac{B^{2}}{μ_{0}}$ will be equal to the dimensions of energy per unit volume.
$[\frac{B^{2}}{μ_{0}}] = [u] = \frac{[{ML}^{2} T^{- 2}]}{[L^{3}]} = [{ML}^{- 1} T^{- 2}]$

Asked in: JEE Main 2022 (28 Jul Shift 1)

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