\(r=\frac{20}{2}\, c m=0.1\, \mathrm{m}\)
\(\mathrm{B}_{\mathrm{net}}=\mathrm{B}_{1}+\mathrm{B}_{2}+\mathrm{B}_{\mathrm{H}}\)
\(B_{n e t}=\frac{\mu_{0}}{4 \pi} \frac{\left(M_{1}+M_{2}\right)}{r^{3}}+B_{H}\)
\(=\frac{10^{-7}(1.2+1)}{(0.1)^{3}}+3.6 \times 10^{-5}\)
\(=2.56 \times 10^{-4}\, \mathrm{wb} / \mathrm{m}^{2}\)