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Using mass (M), length (L), time (T) and...

Using mass (M), length (L), time (T) and current (A) as fundamental quantities, the dimension of permittivity is:

A

`[M^(-1)L T^(-2)A]`

B

`[ML^(-2) T^(-2)A^(-1)]`

C

`[ML T^(-2)A ^(-2)]`

D

`[M L T^(-1) A ^(-1)]`

Text Solution

Verified by Experts

The force per unit length between two parallel current carrying conductors is given by
`F/l =(mu_(0))/(2pi) ((I_(1)I_(2))/(r))`
`therefore [mu_(0)] =([F] [r])/([I_(1)][I_(2)][l])=([M^(1) L^(2)T^(-2) ] [L^(1)])/([A^(1)] [A^(1)][L^(1)])`
`[mu_(0)] =[M^(1)l^(1) T^(-2)A^(-2)]`
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