Q73 / 90 · JEE Main 2022 (28 Jun Shift 2) · Physics

Two parallel, long wires are kept $0.20 m$ apart in vacuum, each carrying current of $x A$ in the same direction. If the force of attraction per meter of each wire is $2 \times 10^{- 6} N$ , then the value of $x$ is approximately

Solution

Force per unit length between two parallel wires is $\frac{F}{l} = \frac{μ_{0} i_{1} i_{2}}{2 π d}$

Here, $i_{1} = i_{2} = i$ and $\frac{F}{l} = 2 \times 10^{- 6} N$

Then, $2 \times 10^{- 6} = \frac{4 π^{2} \times 10^{- 7} i^{2}}{2 π \times 0.2}$

$\Rightarrow i^{2} = \sqrt{2} \Rightarrow i = 1.4 A$

Hence, $x = 1.4$ .

Hence, the correct answer is option: 'C'.