Answer
(a) $I = 20~A$
(b) $r = 1.6~mm$
Work Step by Step
(a) We can find the current:
$B = \frac{\mu_0~I}{2~r}$
$I = \frac{2~r~B}{\mu_0}$
$I = \frac{(2)~(5.0\times 10^{-3}~m)(2.5\times 10^{-3}~T)}{4\pi\times 10^{-7}~T~m/A}$
$I = 20~A$
(b) We can find the distance $r$:
$B = \frac{\mu_0~I}{2~\pi~r}$
$r = \frac{\mu_0~I}{2~\pi~B}$
$r = \frac{(4\pi\times 10^{-7}~T~m/A)(20~A)}{(2~\pi)~(2.5\times 10^{-3}~T)}$
$r = 16\times 10^{-4}~m$
$r = 1.6~mm$
