Let `[in_(0)]` denote the dimensional formula of the permittivity of vacuum. If `M = mass, L = length, T = time and A = elctric current`, then :A. `[In_0]= [M^(-1) l^(-3) T^2 I]`B. `[in_0] = [M^(-1) L^(-3) T^4 I^2]`C. `[mu_0] = [MLT^(-2) I^(-2)]`D. `[mu_0] = [ML^2 T^(-2)I]`

Let `[in_(0)]` denote the dimensional formula of the permittivity of v

1.	Let `[in_(0)]` denote the dimensional formula of the permittivity of vacuum. If `M = mass, L = length, T = time and A = elctric current`, then :A. `[In_0]= [M^(-1) l^(-3) T^2 I]`B. `[in_0] = [M^(-1) L^(-3) T^4 I^2]`C. `[mu_0] = [MLT^(-2) I^(-2)]`D. `[mu_0] = [ML^2 T^(-2)I]`
Answer» Correct Answer - (b,c) As `F = (1)/(4 pi in_0) (q^2)/(r^2) or in_0 = (1)/(4pi) (q^2)/(r^2 F)` `epsilon_0 = (I^2 T^2)/(L^2 MLT^(-2)) = M^(-1) L^(-3)T^4 I^2` As `B = (mu_0)/(4 pi) (I dl sin theta)/(r^2)` `(F)/(qupsilon) =(mu_0)/(4pi) xx (I din sin theta)/(r^2) [ :. F =B q upsilon sin theta]` `mu_0 =(Fr^2)/(q upsilon dl) = (MLT^(-2).L^2)/(IT.LT^(-1). I.L) = MLT^(-2) I^(-2)` Choices (b) and (c ) are correct.