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

Answer

B

Explanation

Magnetic field at the centre of single turn loop `B=(mu_(0))/(4pi)(2pil)/(r)`, magnetic field at the centre of n turn loop
`B_(n)=[(mu_(0))/(4pi)(2piI)/(r//n)]xxn`
`implies B_(n)=n^(2)B`