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

Options

A. `1.4 xx 10^(-4)` N towards the right
B. `1.4xx10^(-4)` N towards the left
C. `2.6 xx 10^(-4)` N to the right
D. `2.6 xx 10^(-4)`to the left

Answer

1

Explanation

Force on wire Q due to R,
`F_(R)=10^(-7)xx 2xx20xx10/(2xx10^(-2)) xx(10xx10^(-2))=2xx10^(-4) m` (Repulsive)
Force on wire Q due to P,
`F_(P)=10^(-7)xx2xx 10xx30/(10xx10^(-2)) xx(10xx10^(-2))=0.6 xx 10^(-4)N` (Repulsive)
Hence net force `F_(net)=F_(R)-F_(P) =2 xx 10^(-4)-0.6 xx 10^(-4) = 1.4 xx 10^(-4)` N (towards right i.e. in the direction of `vec F_(R)`)