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Let [in(0)] denote the dimensional formu...

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]`

Text Solution

Verified by Experts

The correct Answer is:
(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.
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