<Dd> <Dl> <Dd> d B = α B I d l × r ^ r 2, (\ displaystyle d \ mathbf (B) = \ alpha _ (\ rm (B)) (\ frac (Id \ mathbf (l) \ times \ mathbf (\ hat (r))) (r ^ (2))) \;,) where r and r ^ (\ displaystyle \ mathbf (\ hat (r))) are the length and the unit vector in the direction of vector r respectively . </Dd> </Dl> </Dd> <Dl> <Dd> d B = α B I d l × r ^ r 2, (\ displaystyle d \ mathbf (B) = \ alpha _ (\ rm (B)) (\ frac (Id \ mathbf (l) \ times \ mathbf (\ hat (r))) (r ^ (2))) \;,) where r and r ^ (\ displaystyle \ mathbf (\ hat (r))) are the length and the unit vector in the direction of vector r respectively . </Dd> </Dl> <Dd> d B = α B I d l × r ^ r 2, (\ displaystyle d \ mathbf (B) = \ alpha _ (\ rm (B)) (\ frac (Id \ mathbf (l) \ times \ mathbf (\ hat (r))) (r ^ (2))) \;,) where r and r ^ (\ displaystyle \ mathbf (\ hat (r))) are the length and the unit vector in the direction of vector r respectively . </Dd> <P> These two laws can be used to derive Ampère's force law above, resulting in the relationship: k A = α L ⋅ α B (\ displaystyle k_ (\ rm (A)) = \ alpha _ (\ rm (L)) \ cdot \ alpha _ (\ rm (B)) \;). Therefore, if the unit of charge is based on the Ampère's force law such that k A = 1 (\ displaystyle k_ (\ rm (A)) = 1), it is natural to derive the unit of magnetic field by setting α L = α B = 1 (\ displaystyle \ alpha _ (\ rm (L)) = \ alpha _ (\ rm (B)) = 1 \;). However, if it is not the case, a choice has to be made as to which of the two laws above is a more convenient basis for deriving the unit of magnetic field . </P>

What is the full form of cgs in physics