Answer
$F=(-2.50mN)j^{\wedge}+(0.750mN)k^{\wedge}$
Work Step by Step
$F=IL\times B$
We plug in the known values to obtain:
$F=(0.500A) (0.500i^{\wedge}\times(0.003Tj^{\wedge}+0.010Tk^{\wedge}))$
We know that $i^{\wedge}\times j^{\wedge}=k^{\wedge}$ and $i^{\wedge}\times k^{\wedge}=-j^{\wedge}$
So,
$F=(0.500)(0.0015k^{\wedge}+0.005(-j^{\wedge}))$
$F=0.00075k^{\wedge}-0.0025j^{\wedge}$
$F=(-2.50mN)j^{\wedge}+(0.750mN)k^{\wedge}$
