A circular coil of length L and radius r carries a current I. It provides a magnetic field of induction B at its center. Now the coil is bent sharply so as from a circular coil of n turns of rqual radii. The new induction at the center of this coil is

Options

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

Answer

B

Explanation

`L=2pir=2pinr' therefore r=nr' B'/B=(2pinI)/(r') xx r/(2piI)=(nr)/(r')=(n(nr'))/(r')=n^(2)therefore B'-n^(2)B`