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

Answer

C

Explanation

`L=2pir=2pir' therefore r=nr' therefore B/B=(mu_0ni)/(2r') xx (2r)/(mu_0I)=(nr)/(r')=(n nr')/(r')=n^(2)`