Units & Dimensions

JEE Main level, single correct, medium

120 of 149

Question 120 of 149

Given, BB is magnetic field induction, and μ0{\mu }_{0} is the magnetic permeability of vacuum. The dimension of B22μ0\dfrac{{B}^{2}}{2{\mu }_{0}} is:

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Answer: (D) ML−1T−2{ML}^{-1}{T}^{-2}

Energy density in magnetic filed =energyvolume=B22μ0=\dfrac{energy}{volume}=\dfrac{{B}^{2}}{2{\mu }_{0}}
=Force×displacement(displacement)3=MLT−2.LL3=ML−1T−2=\dfrac{Force\times displacement}{{\left(displacement\right)}^{3}}=\dfrac{ML{T}^{-2}.L}{{L}^{3}}={ML}^{-1}{T}^{-2}

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