Correct Answer - D
(d) Case-I: `x gt R/2`
`|B|=0`
Case-II: `R/2lexleR`
`intvecB.vec(dl)=mu_0I`
`|B|2pix=mu_0[pix^2-pi(R/2)^2]J`
`|B|=(mu_0J)/(2x)(x^2-(R^2)/4)`
Case III: `xgeR`
`intvecB.vec(dl)=mu_0I`
`|B|2pix=mu_0[piR^2-pi(R/2)^2]J`
`|B|=(mu_0J)/(2x) 3/4 R^2`
`|B|=(3mu_0JR^2)/(8x)`
so