NTA Abhyas JEE Main2020PhysicsUnits and DimensionsPractice
The dimension of B 2 2 μ 0 , where B is magnetic filed and μ 0 is the magnetic permeability of vacuum, is:
Options
- AM L T - 2
- BM L 2 T - 1
- CM L 2 T - 2
- DM L - 1 T - 2
Correct answer
D. M L - 1 T - 2
Step-by-step solution
Energy density in magnetic filed = B 2 2 μ 0 = F o r c e × d i s p l a c e m e n t d i s p l a c e m e n t 3 = M L T - 2 . L L 3 = M L - 1 T - 2