A long wire carries a steady curent . It is bent into a circle of one turn and the magnetic field at the centre of the coil is `B`. It is then bent into a circular loop of `n` turns. The magnetic field at the centre of the coil will be

Options

A. `n^(2)B`
B. `2 nB`
C. `2n^(2)B`
D. `nB`

Answer

A

Explanation

The field at the centre of a coil is
`B =(mu_(0))/(4pi) ((2pii)/(r))or (mu_(o)i)/(2r)`
In the first case, `l=2pir`
In the second cse, `l =2pi n r'`
`therefore 2pir = 2 pi nr ' therefore r =nr'`
`(B')/(B) =(mu_(0)i xxn)/(2r') xx(2r )/(mu_(0)i)=(nr)/(r ') =( n nr')/(r') =n^(2)`
`therefore B' =n^(2) B`