Question 194: Units , Dimensions and Error Analysis

Question

The dimension of ${{{B^2}} \over {2{\mu _0}}}$, where B is magnetic field and ${{\mu _0}}$
 is the magnetic permeability of vacuum,
is :

Accepted Answer

As ${{{B^2}} \over {2{\mu _0}}}$ = Energy per unit volume
$ 	herefore $ Dimension = ${{M{L^2}{T^{ - 2}}} \over {{L^3}}}$ = ML
-1
T
Ã¢ÂÂ2

Answer

ML
2
T
–2

Answer

MLT
–2

Answer

ML
-1
T
–2

Answer

ML
2
T
–1

Units , Dimensions and Error Analysis