Solution
Method of Effective Length:
For a current carrying wire in a uniform magnetic field, the force is given by $\vec{F} = I (\vec{L}_{eff} \times \vec{B})$, where $\vec{L}_{eff}$ is the vector connecting the entry and exit points in that region.
Region 1 (Left of A1A2):
The wire enters at point $A_2$ and exits at $A_1$ (or vice versa). The effective length vector connects $A_2$ to $A_1$. Based on the grid, the distance $A_1 A_2$ corresponds to 8 unit cell lengths ($8l$). Let’s assume current flows upwards. $\vec{L}_{left} = 8l \hat{j}$. The field is $\vec{B} = B \hat{k}$ (Out).
$$ \vec{F}_{left} = I (8l \hat{j} \times B \hat{k}) = 8 I l B \hat{i} $$Region 2 (Right of A1A2):
For the closed loop, the return current flows down the right side. The effective length vector is from $A_1$ to $A_2$, so $\vec{L}_{right} = -8l \hat{j}$. The field is $\vec{B} = -B \hat{k}$ (In).
$$ \vec{F}_{right} = I (-8l \hat{j} \times (-B \hat{k})) = 8 I l B (\hat{j} \times \hat{k}) = 8 I l B \hat{i} $$Total Force:
The total force is the sum of forces on both halves. Since both contributions point in the same direction:
$$ \vec{F}_{total} = 8 I l B \hat{i} + 8 I l B \hat{i} = 16 I l B \hat{i} $$