Three long, straight and parallel wires are arranged as shown in Fig. The forces experienced by 10 cm length of wire Q is

Options

A. `1.4xx10^(-4)N` towards right
B. `1.4xx10^(-4)N` towards left
C. `2.6xx10^(-4)N` towards right
D. `2.6xx10^(-4)N` towards left

Answer

A

Explanation

Force on wire `Q` due to wire `P` is
`F_(p)=10^(-7)xx(2xx30xx10)/(0.1)=6xx10^(-5)N` (Towards left)
Force on wire `Q` due to wire `R` is
` F_(R)=10^(-7)xx(2xx20xx10)/(0.02)xx0.1=20xx10^(-5)N` (Towards right)
Hence `F_(net)=F_(r)-F_(p)=14xx10^(-5)N=1.4xx10^(-4) N` (Towards right)