Phy problem...electricity4

Use Ampere's Law,

B.(2.pi.r) = (uo).I
where uo is the permeability of free space
I is the enclosed current
hence, B = (uo).I/[2.pi.r]

But I = J.(pi.r^2) = (Jo).(r/a).(pi.r^2) = (Jo).(pi).(r^3/a)
B = [(uo)(Jo).(pi).(r^3/a)]/[2.pi.r] = (uo).(Jo)(r^2)/2a = 0.06283(r^2)

(a) When r = 0, B = 0 T

(b) B = 0.06283 x (3.1x10^-3/2)^2 T = 1.509 x 10^-7 T

(c) B = 0.06283 x (3.1x10^-3)^2 T = 6.04 x 10^-7 T