Question 179: 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

ML2T–2

Answer

MLT–2

Answer

ML-1T–2

Answer

ML2T–1

Units , Dimensions and Error Analysis