Question

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

• A. $1.4\times {10}^{-4}N$ towards the right
• B. $1.4\times {10}^{-4}N$ towards the left
• C. $2.6\times {10}^{-4}N$ to the right
• D. $2.6\times {10}^{-4}N$ to the left

Answer 1

Answer: (A)

$1.4\times {10}^{-4}N$ towards the right

Force on wire $Q$ due to wire $P$ is

$F_P={10}^{-7}\times \frac{2\times 30\times 10}{0.1}\times 0.1=6\times {10}^{-5}N$ (Towards left)

Force on wire $Q$ due to wire $R$ is

$F_R={10}^{-7}\times \frac{2\times 20\times 10}{0.02}\times 0.1=20\times {10}^{-5}N$ (Towards right)

Hence $F_{net}=F_R-F_P$

$=14\times {10}^{-5}N$

$=1.4\times {10}^{-4}N$ (Towards right)