Question

Three long straight and parallel wires, carrying are arranged as shown figure. The force experienced by the conductor 'B' of length 25 cm is

• A. $4\times {10}^{-4}N$ from left to right
• B. $4\times {10}^{-5}N$ towards left side
• C. $2\times {10}^{-4}N$ from left to right
• D. $2\times {10}^{-4}N$ from right to left

Answer 1

The correct option is B $4\times {10}^{-5}N$ towards left side

Length of the wire on which force is experienced$,I=25cm$

As seen in the fig$.$ the direction of current in $C$ and $D$ wires are in opposite direction

So$,$ force of repulsion F will act on $C$ towards right side$.$

Similarly$,$ due to $G$ it will be in left hand side so resultant force on $C$ due to wires $D$ and $G$ will be $F.$

Thus$,$

$F=F_1-F_2$

$=\frac{{\mu }_0{I}_1{I}_{2}l}{2\pi r_1}-\frac{{\mu }_0{I}_1{I}_{3}l}{2\pi r_2}$

$=\frac{{\mu }_0{I}_{1}l}{2\pi r}\left[\frac{I_2}{r_1}-\frac{I_3}{r_2}\right]$

$=\frac{4\pi \times {10}^{-7}\times 10\times 25}{2\pi }\left[\frac{30}{0.03}-\frac{20}{0.10}\right]$

$=5\times {10}^{-7}\times 80$

$=4\times {10}^{-5}N$

Hence,

option $\left(B\right)$ is correct answer.