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

Answer

D

Explanation

Since `l = 2pi R = n(2pi r) implies r = (R)/(n)`
Force one turn `B = (mu_(0) i)/(2R)` and
For `n` turns `B' = (mu_(0) ni)/(2r) implies B' = (mu_(0) n^(2) i)/(2R) = n^(2) B`