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Let [epsilon(0)] denote the dimensional ...

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`,

A

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

B

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

C

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

D

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

Text Solution

Verified by Experts

The correct Answer is:
B, C
By defition `F = (Q_(1) Q_(2))/(( 4 pi epsilon_(0)) r^(2)) and (F)/(l) = ( mu_(0) I_(1) I_(2))/( 2 pi L)`
Hence , `[epsilon _(0)] = ([Q^(2)])/([ F] [r ^(2)]) = ( I^(2) T^(2))/( [ MLT^(-2)] [L^(2)]) = [M^(-1)L^(-3)T^(4) I^(2)]`
`[mu_(0)] = ([F])/([I^(2)]) = ([MLT^(-2)])/([I^(2)]) = [MLT^(-2)I^(-2)]`
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