Question

Through two parallel wires A and B, $10\mathrm{A}$ and $2\mathrm{A}$ of currents are passed respectively in opposite directions. If the wire A is infinitely long and the length of the wire B is $2\mathrm{m}$, then force on the conductor B, which is situated at a $10\mathrm{cm}$ distance from A, will be

Answer 1

Step 1: Given

The value of the currents in the wire is

$$\begin{gathered} {\mathrm{I}}_1=10\mathrm{A} \\ {\mathrm{I}}_2=2\mathrm{A} \end{gathered}$$

The length of the wire is $\mathrm{l}=2\mathrm{m}$

The value of the distance between the wires is,

$$\begin{gathered} \mathrm{r}=10\mathrm{cm} \\ =0.1\mathrm{m} \end{gathered}$$

Step 2: To find

We have to find the force on conductor B, which is situated at a $10\mathrm{cm}$ distance from A.

Step 3: Calculation

1. Ampere's circuital law is used to calculate the force between two long straight and parallel wires distanced by 'r'.
2. Whenever the current flows in the opposite direction, this same net force on this cable is outwards. The net force is directed outward.

The diagrammatical representation of the given case is,

The force on the wire B due to the current in wire A is,

$\mathrm{F}=\frac{{\mathrm{\mu }}_0}{4\mathrm{\pi }}\frac{2{\mathrm{I}}_1{\mathrm{I}}_2}{\mathrm{r}}\times \mathrm{l}$

By substituting the values, we get
$$\begin{gathered} =\frac{{10}^{-7}\times 2\times 10\times 2}{0.1}\times 2 \\ =8\times {10}^{-5}\mathrm{N} \end{gathered}$$

Therefore, the force on the conductor B is $8\times {10}^{-5}\mathrm{N}$.