Units & Dimensions

JEE Main level, single correct, medium

85 of 149

Question 85 of 149

The dimensions of (B2μ0)\left(\dfrac{{B}^{2}}{{\mu }_{0}}\right) will be
(if μ0{\mu }_{0}: permeability of free space and B:B:magnetic field)

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Answer: (C) [ML−1T−2]\left[{ML}^{-1}{T}^{-2}\right]

Energy density in magnetic field is given by u=B22μ0u=\dfrac{{B}^{2}}{2{\mu }_{0}}. Therefore, dimensions of B2μ0\dfrac{{B}^{2}}{{\mu }_{0}} will be equal to the dimensions of energy per unit volume.
[B2μ0]=[u]=[ML2T−2][L3]=[ML−1T−2]\left[\dfrac{{B}^{2}}{{\mu }_{0}}\right]=\left[u\right]=\dfrac{\left[{ML}^{2}{T}^{-2}\right]}{\left[{L}^{3}\right]}=\left[{ML}^{-1}{T}^{-2}\right]

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