Using mass `(M)` , length `(L)` , time `(T)` , and electric current `(A)` as fundamental quantities , the dimensions of permitivity will be

Options

A. `epsilon_0=[M^-1L^-3T^2I]`
B. `epsilon_0=[M^-1L^-3T^4I^2]`
C. `mu_0=[MLT^-2I^-2]`
D. `mu_0=[ML^2T^-1I]`

Answer

(b,c)

Explanation

Force between two charges `q_1,q_2` certain distance
apart in vacuum is
`F=(q_1q_2)/(4pi in_0 r^2) or epsilon_0=(q_1q_2)/(4Fr^2)`
Dimensional formula of `epsilon_0 =((IT)^2)/((MLT^-2)(L^2))`
`=[M^-1L^-3T^4I^2]`
Force of attraction acting per unit length of two
long linear parallel conductors carrying currents
in the same direction is
`F=(mu_0)/(4pi) (2I_1I_2)/r or mu_0=(4pirF)/(2I_1I_2)`
Dimensional formula of
`mu_0=(LxxMLT^-2//L)/(IxxI)`
`=[M^1L^1T^-2I^-2]`