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