$${B = \frac{(4\pi \times 10^{-7}\, \mathrm{T \cdot m/A})}{4 \pi} \frac{(10 \times 10^{-6}\, \mathrm{A})(1 \times 10^{-3}\, \mathrm{m})}{(2 \times 10^{-2}\, \mathrm{m})^2} \sin 90^{\circ} \\ = 2,5 \times 10^{-12}\, \mathrm{T} \approx 3\, \mathrm{pT}}$$