A current carriyng straight conductor OA lies along x-axis. In that region a magnetic field of induction B exists whose direction is inward and magnitude of ‘B’ increases as we are moving along positive x-axis. If C be the mid point of the rod OA, then select the correct option.

If we consider a small element $∆\mathrm{l}$ of the rod, then magnetic force acting on the element is $∆\mathrm{F}=\mathrm{i}.∆\mathrm{l}.\mathrm{B}$. Where B is the magnetic induction at the location of $∆\mathrm{l}$. If B is uniform , the resultant force acts at C. Since B is increasing as we are moving along positive x-axis the line of action of resultant  force shifts to the right. Hence option -(2) is correct.