Explanation

(d) Magnetic field at point $P$ due to $i_{1},\ B_{1} = \frac{\mu_{0}i_{1}}{2\pi x}\ $ upward perpendicular to plane paper. Therefore, magnetic force on element $P$
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$F_{P} = B_{1}i_{2}dx = \frac{\mu_{0}i_{1}i_{2}}{2\pi x}\ dx,$ along $x$-axis
Magnetic field at point $Q$ due to $i_{1}$
$B_{2} =$ $\frac{\mu_{0}i_{1}}{2\pi x},$ downward perpendicular to plane of paper.
$\ F_{Q} = B_{2}i_{x}dx = \frac{\mu_{0}i_{1}i_{2}}{2\pi x}\ dx,$ along (+) x-axis
So, wire $i_{2}$ will be anticlockwise.