A long wire carrying a steady current is bent into a circular loop of one turn. The magnetic field at the centre of the loop is B. it is then bent into a circular coil of n turns. The magnetic field at the centre of this coil of n turns will be

Options

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

Answer

B

Explanation

The magnetic field at the centre of circular coil is
`B=(mu_(0)1)/(2R)`
Where r=radius of circle `=1/(2pi) [:'I=2pir]`
`:.B=(mu_(0)I)/2xx(2pi)/I=(mu_(0)//pi)/1`….i
When wire of length l bents into a circular loops of n turns, then `1=nxx2pir'impliesr'=1/(nxx2pi)`
Thus new magetic field `X_(l)=100sqrt(3)Omega=(2pifL)=(mu_(0)Ipi)/1xxn^(2)=n^(2)B` [ From Eq. (i)]