A wire of some length is bent in the form of a ring of diameter `2a` having self inductance `L`, then `L` will depend upon `a` as:

Options

A. `a^(0)`
B. `a^(-1)`
C. `a^(2)`
D. `a^(-2)`

Answer

B

Explanation

`phi=NBA`
`LI=N.(mu_(0)I)/(2a)xxpia^(2)N`
`Now, `Nxx2pia=l`
N=(l)/(2pia)` so, `LI=(l)/(2pia)xx(mu_(0)I)/(2a)xxpia^(2)xx(l)/(2pia)`
`Lpropa^(-1)`