The dimensions of , where B is magnetic field and μ0 is the magnetic permeability of vacuum, is

Engineering

Physics

Dimensions

Question

The dimensions of $\frac{B^{2}}{2 μ_{0}}$ , where B is magnetic field and μ₀ is the magnetic permeability of vacuum, is

ML²T^–2

ML²T^–1

MLT^–2

ML^–1T^–2

College PredictorLive

Know your College Admission Chances Based on your Rank/Percentile, Category and Home State.

Get your JEE Main Personalised Report with Top Predicted Colleges in JoSA

Energy density in magnetic field $= \frac{B^{2}}{2 μ_{0}}$

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