Three long straight and parallel wires are arranged as shown in figure. The force experienced by 10 cm length of wire ‘Q’ is

Wire Q and R will repel and wire Q and P will also repel. As shown in figure, net force can be given as,  Fnet=FQR-FQP ⇒Fnet=μo2πiRiQLdQR-μo2πiQiPLdQP ⇒Fnet=4π×10−7×20×10×10×10−22π×2×10−2−4π×10−7×10×30×10×10−22π×10×10−2 ⇒Fnet=4π×10−7×10×10−22π×10−2[100−30]= 20×10−7×70=1400×10−7= 1.4×10−4N towards right