The expression for the magnetic force on a current lines is given by:
[tex]F=\text{ilBsin}\theta[/tex]where,
i: current = 10 A
B: magnitude of the magnetic field = 0.2T
l: length of the line
θ: angle between direction of the line and the magnetic field = 90 degrees.
You can notice that the direction of the magnetic field and direction of the conduction are perpendicular, then, you have for F:
[tex]F=i\cdot l\cdot B\sin 90=i\cdot l\cdot B[/tex]For the segment BA you obtain:
[tex]F_{BA}=(10A)(0.08m)(0.2T)=0.16N[/tex]and for segment AG:
[tex]F_{AG}=(10A)(0.06m)(0.2T)=0.12T[/tex]If you consider that segment BA lies on the y axis of a coordinate system and segment AG lies on x-axis (in this case BA and AG form a right angle), then, the total force on BAG is:
[tex]\vec{F}=0.12T\hat{i}+0.16T\hat{j}[/tex]and its magnitude is:
[tex]F=\sqrt[]{(0.12T)^2+(0.16T)^2}=0.2N[/tex]