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

Two parallel wires are separated by $$d = 0.20$$ m, each carrying current $$x$$ A. The force of attraction per meter is $$F/L = 2 \times 10^{-6}$$ N m$$^{-1}$$.

Write the formula for force per unit length between two parallel current-carrying wires.

$$\frac{F}{L} = \frac{\mu_0 I_1 I_2}{2\pi d}$$

Since both wires carry the same current $$x$$:

$$\frac{F}{L} = \frac{\mu_0 x^2}{2\pi d}$$

Substitute the known values.

$$2 \times 10^{-6} = \frac{4\pi \times 10^{-7} \times x^2}{2\pi \times 0.20}$$

$$2 \times 10^{-6} = \frac{2 \times 10^{-7} \times x^2}{0.20}$$

$$2 \times 10^{-6} = 10^{-6} \times x^2$$

Solve for $$x$$.

$$x^2 = 2$$

$$x = \sqrt{2} \approx 1.4 \text{ A}$$

The correct answer is Option C.