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

Answer

B

Explanation

Let l be the length of the wire. Magnetic field at the centre of the loop is
`B = (mu_0l)/(2R)`
`therefore B = (mu_0 piI)/(l) " " (therefore l = 2pi R) .....(i)`
`B. = (mu_0nI)/(2r) = (mu_0 nl)/(2(l/(2npi)))`
`B. = (mu_0 n^2 pi l)/(l)`
From eqns. (i) and (ii), we get B. = `n^2 B`