Let `varepsilon_(0)` denote 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, then the dimensional formula of the

Options

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

Answer

B, C

Explanation

`F=(1)/(4pi in_(0))(q_(1)q_(2))/r^(2) rArr MLT^(-2)=[1/in_(0)](A^(2)T^(2))/L^(2)`
`rArr [1/in_(0)]=M^(1)L^(3)A^(-2)T^(-4) rArr [in_(0)]=M^(-1)L^(-3)A^(+2)T^(4)`
`F/L=mu_(0)/(4pi)(i_(1)i_(2))/(r)`
`[ML^(-2)]=[(mu_(0)A^(2))/L]rArr [mu_(0)]=MLA^(-2)T^(-2)`