Correct Answer - B
a.
`a=1/2((2pir)/n)=(pir)/n`
`x=acotpi/n=((pir)/n)cot(pi/n)`
`B=n[mu_0/(4pi)i/x(sinpi/n+sinpi/n)]`
`=n[mu_0/(4pi)i/x(sinpi/n+sinpi/n)]`
`=n[mu_0/(4pi)i/((pir/n)cotpi/n)(2sinpi/n)]`
`=(mu_in^2sin(pi/n)tan(pi/n))/(2pir)`
b. the above calculated magnetic field can be written is
`B=(mu_0i)/(2r) (((sinpi/n)/(pi/n))^2)/(cospi/n)`
`As nrarroopi/nrarr0`
`Hence, lim_(pi/nrarr0) ((sinpi/n)/(pi/n))rarr1`
`and lim_(pi/nrarr0) (cotpi/n)rarr1`
`or Brarr(mu_0i)/(2r)`