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Let [epsi(0)] denote the dimensional for...

Let `[epsi_(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 :

A

`[epsi_(0)]=[M^(-1)L^(-3)T^(2)I]`

B

`[epsi_(0)]=[M^(-1)L^(-3)T^(4)I^(2)]`

C

`[mu_(0)]=[MLT^(-2)I^(-2)]`

D

`[mu_(0)]=[ML^(2)T^(-1)I]`

Text Solution

Verified by Experts

The correct Answer is:
b, c
Since, `F=1/(4pi epsi_(0))(q_(1)q_(2))/r^(2)`
Hence, `epsi_(0)=((q_(1)q_(2))/(4pi Fr^(2)))`
and `q=I xxt`
Hence, `epsi_(0) rarr [M^(-1)L^(-3)T^(4)I^(2)]`
`B=mu_(0)/(2pi) I/r`
`mu_(0)=((Bxx2pixxr)/I)`
Hence, `mu_(0) rarr [ML^(2)T^(-1)]`
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