Answer
30 A
Work Step by Step
Model the situation as a long straight wire. Use equation 20–6.
$$B_{max}=\frac{\mu_oI}{2\pi r}$$
$$I=\frac{2\pi r B_{max}}{\mu_o }$$
$$=\frac{2\pi(12\times10^{-2}m) (0.50\times10^{-4}T)}{4\pi \times10^{-7} T\cdot m/A }$$
$$=30A$$
