Three long straight parallel wires, carrying current, are arranged as shown in figure. The force experienced by 25 length of wire C is :

The magnetic field due to wire D at wire C is :BD=μ04π2Ir=10−π×2×300.03=2×10−4 Twhich is directed into the page.The magnetic field due to wire G at C is,BG=10−7×2×200.02=2×10−4Twhich is directed out of page.Therefore, the field at the position of wire C is :B=BD−BG=2×10−4−2×10−4 = 0The force on 25 cm of wire C is :F=BIlsin⁡θ=0