Answer
7.5 A
Work Step by Step
The experiment is to be accurate to $\pm3.0\%$, so the magnetic field due to the current should be, at most, $3.0\%$ of the Earth’s magnetic field.
Model the situation as a long straight wire. Use equation 20–6.
$$B_{max}=\frac{\mu_oI}{2\pi r}$$
$$I=\frac{2\pi r B_{max}}{\mu_o }$$
$$=\frac{2\pi(1.00m)(0.030) (0.50\times10^{-4}T)}{4\pi \times10^{-7} T\cdot m/A }$$
$$=7.5A$$
