Chapter 29 - The Magnetic Field - Exercises and Problems - Page 830: 10

$I = 4.0\times 10^{-8}~A$

We can find the peak current: $B = \frac{\mu_0~I}{2~\pi~r}$ $I = \frac{2~\pi~r~B}{\mu_0}$ $I = \frac{(2~\pi)~(1.0\times 10^{-3}~m)(8.0\times 10^{-12}~T)}{4\pi\times 10^{-7}~T~m/A}$ $I = 4.0\times 10^{-8}~A$