


Reason : $I_1 = I_2$ implies that the fields due to the current $I_1$ and $I_2$ will be balanced.
$(A)$ If $\vec{B}$ is along $\hat{z}, F \propto(L+R)$
$(B)$ If $\overrightarrow{ B }$ is along $\hat{ x }, F =0$
$(C)$ If $\vec{B}$ is along $\hat{y}, F \propto(L+R)$
$(D)$ If $\overrightarrow{ B }$ is along $\hat{ z }, F =0$