Question

Equal currents are flowing in four infinitely long wires. Distance between any two adjacent wires is same, then

$\begin{array}{cc}\text{Column I}& \text{Column II}\\ 1)\text{force on wire}1& \text{p) inwards}\\ 2)\text{force on wire}2& \text{q) leftwards}\\ 3)\text{force on wire}3& \text{r) rightwards}\\ 4)\text{force on wire}4& \text{s) zero}\end{array}$

• A. $1-\text{q},2-\text{r},3-\text{q},4-\text{r}$
• B. $1-\text{p},2-\text{q},3-\text{p},4-\text{r}$
• C. $1-\text{q},2-\text{p},3-\text{q},4-\text{p}$
• D. $1-\text{p},2-\text{r},3-\text{p},4-\text{r}$

Answer 1

The correct option is A $1-\text{q},2-\text{r},3-\text{q},4-\text{r}$

Let the distance between each wire be $a$.

For wire-$1$, the magnetic field due to wire-$2$, wire-$3$ and wire-$4$ respectively is given by,


$B_2=\frac{{\mu }_{0}I}{2\pi a},B_3=\frac{{\mu }_{0}I}{4\pi a},B_4=\frac{{\mu }_{0}I}{6\pi a}$

$|{\overrightarrow{B}}_{net}|=\frac{{\mu }_{0}I}{2\pi a}-\frac{{\mu }_{0}I}{4\pi a}+\frac{{\mu }_{0}I}{6\pi a}=\frac{5{\mu }_{0}I}{12\pi a}$

This net field is directed normally inwards.

$\therefore$ force on wire-$\left(1\right)$ is leftwards.

Wire (2)


$|\overrightarrow{B}|=\frac{{\mu }_{0}I}{2\pi a}+\frac{{\mu }_{0}I}{4\pi a}-\frac{{\mu }_{0}I}{2\pi a}=\frac{{\mu }_{0}I}{4\pi a}$

This field is directed normally inwards.

$\therefore$ force on wire-$\left(2\right)$ will be directed rightwards

For wire-$3$,


$|\overrightarrow{B}|=B_1-B_2+B_4=\frac{{\mu }_{0}I}{4\pi a}$

This field is directed normally inwards.

$\therefore$ force on wire (3) will be directed leftwards

For wire-$4$,


$|\overrightarrow{B}|=B_1+B_3+B_2=\frac{5{\mu }_{0}I}{12\pi a}$

This field is directed normally inwards.

$\therefore$ the force on wire (4) will be directed rightawards