Let `[epsilon_(0)]` denote the dimensional formula of the permittivity of the vacuum, and `[mu_(0)]` that of the permeability of the vacuum. If `M = mass ,L = length, T = time and I = electric current`,

Options

A. `[epsilon_(0)]`= [M^(-1)L^(-3)T^(3)l]`
B. `[epsilon_(0)]= [M^(-1)L^(-3)T^(4)l^(2)]`
C. `[mu_(0)]= [MLT^(-2)l^(-2)]`
D. `[epsilon_(0)]= [M^(-1)L^(-3)T^(3)l]`

Answer

c

Explanation

`[epsilon_(0)] = ([q_(1)][q_(2)])/([F][r^(2)]) = ([IT]^(2))/([MLT^(-2)][L^(2)])= [M^(-1)L^(-1) T^(-4)I^(2)]`
and `[mu_(0)] = (1)/([epsilon_(0)][c^(2)])= (1)/(M^(-1)L^(-3)T^(4)I^(2)][LT^(-1)]^(2)`
`= [MLT^(-2)I^(-2)]`.