- Using mass(M) , length(L) ,time(T) and current (A) as fundamental quantities, the dimension of permeability is (1)M^-1LTA. (2)ML^2T^-2A^-1 (3)MLT^-2A^-2. (4)MLT^-1A^-1
Asked by Kajalmandal64
12th May 2018,
8:26 PM
Answered by Expert
Answer:
Let us consider Coulomb's law : F = q2 /(4πε0×r2) ;
q is charge in Coulomb, r is distance in metre, F is force in Newton and ε0 is permeability.
dimenison of ε0 = dimension of q2/(F×r2) = [Coulomb]2 /[Newton× square metre]................(1)
dimension of Coulomb = dimension of (Current×Time) = [ A T ]
dimension of Newton = [ MLT-2]
substitute the above in (1), we have
dimension of ε0 = [ A2T2 ]/ [MLT-2 × L2] = [ A2 M-1 L-3 T4 ]
Answered by Expert
13th May 2018,
10:21 AM
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