Question

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

• 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

The correct option is A $1.4\times {10}^{-4}N$ towards the right


Force on wire Q due to R ; $F_R={10}^{-7}\times \frac{2\times 20\times 10}{(2\times {10}^{-2})}\times (10\times {10}^{-2})=2\times {10}^{-4}m\left(Repulsive\right)$
Force on wire Q due to P ; $F_P={10}^{-7}\times 2\times \frac{10\times 30}{(10\times {10}^{-2})}\times (10\times {10}^{-2})=0.6\times {10}^{-4}N$
(Repulsive)
Hence net force $F_{net}=F_R-F_P=2\times {10}^{-4}-0.6\times {10}^{-4}=1.4\times {10}^{-4}N$ (towards right i.e. in the direction of $\overrightarrow{F_R}$).