The force on a conducting element of length $$d\overrightarrow l $$, carrying a current I in a magnetic field $$\overrightarrow B $$, is given by d$$\overrightarrow F $$ = I d$$\overrightarrow l $$ $$\times$$ $$\overrightarrow B $$. If the field is uniform, the total force on the conductor is given by

$$\overrightarrow F = \int_a^g {Id\overrightarrow l \times \overrightarrow B = I\left( {\int_a^g {d\overrightarrow l } } \right) \times \overrightarrow B = I\,\overrightarrow {ag} \times \overrightarrow B } $$

Note that $$\overrightarrow B $$ is taken out of the integral sign because it is constant. The vector $$\overrightarrow {ag} = 2(L + R)\widehat x$$. The magnetic forces on the conductor in the given cases are

Case (A) : $$\overrightarrow F = I(2(L + R)\widehat x) \times (B\widehat z)$$

$$ = - 2IB(L + R)\widehat y$$

Case (B) : $$\overrightarrow F = I(2(L + R)\widehat x) \times (B\widehat x) = \overrightarrow 0 $$

Case (C) : $$\overrightarrow F = I(2(L + R)\widehat x) \times (B\widehat y)$$

$$ = 2IB(L + R)\widehat z$$