The forces on a current-carrying wire in a magnetic field is:$F = l(L \times B)$Given:I = 0.5A, $L = 1 \hat{i} \, \mathrm{cm} \quad B = (0.4 \mathrm{mT} \hat{j} + 0.6 \mathrm{mT} \hat{k})$Using cross product:$L \times B = \hat{i} \times (0.4 \hat{j} + 0.6 \hat{k})$Using vector cross-product rules:$\hat{i} \times \hat{j} = \hat{k}, \quad \hat{i} \times \hat{k} = -\hat{j}$$\hat{i} \propto (0.4 \hat{j} + 0.6 \hat{k}) = 0.4 \hat{k} - 0.6 \hat{j}$Multiply by I = 0.5:$\overline{F} = (0.2 \hat{k} - 0.3 \hat{j})$Length was in cm and magnetic field in mT.So, in SI unit $\vec{F} = (0.2 \hat{k} - 0.3 \hat{j}) \times 10^{-5} \, \mathrm{N}$$\vec{F} = (2 \hat{k} - 3 \hat{j}) \, \mu \mathrm{N}$