Answer
(a) $\mu = 3.1\times 10^{-4}~A~m^2$
(b) $B = 5.0\times 10^{-7}~T$
Work Step by Step
(a) We can find the magnetic moment:
$\mu = I~A$
$\mu = I~\pi~r^2$
$\mu = (100~A)~(\pi)~(1.0\times 10^{-3}~m)^2$
$\mu = 3.1\times 10^{-4}~A~m^2$
(b) We can find the on-axis field strength:
$B = \frac{\mu_0~\mu}{2\pi~r^3}$
$B = \frac{(4\pi\times 10^{-7}~T~m/A)(3.1\times 10^{-4}~A~m^2)}{(2\pi)~(0.050~m)^3}$
$B = 5.0\times 10^{-7}~T$
