Three long, straight and parallel wires C, D and G carrying currents are arranged as shown in Fig. The force experienced by a 25 cm length of wire C is
4×10−4N
The magnetic field due to wire D at wire C is
BD=(μ04π)2Ir=10−7×2×300.03=2×10−4T
which is directed into the page.
Similarly, the field due to wire G at C is
BG=10−7×2×200.1=0.4×10−4 T
which is directed out of the page.
Therefore, the field at the position of the wire C is
B=BD−BG=2×10−4−0.4×10−4=1.6×10−4T
and is directed into the page.
The force on 25 cm of wire C is
F = BIl 90∘=1.6×10−4×10×0.25
=4×10−4N