The dimensions of , where B is magnetic field and is the magnetic permeability of vacuum, is

JEE MainPhysicsEasyMagnetism

The dimensions of $\frac{B^{2}}{2\mu_{0}}$ , where B is magnetic
field and $\mu_{0}$ is the magnetic permeability of vacuum, is

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Energy density in magnetic field

= \frac{B^{2}}{2\mu_{0}}

= \frac{\ Force\ \times \ displacement\ }{(\ displacement\ )^{3}} = \frac{{MLT}^{- 2}L}{L^{3}} = {ML}^{- 1}T^{- 2}

Source: Grass-G / P-13 Physics paper 4 Nov.docx

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