1. Use force between two parallel current-carrying wires

For two parallel wires separated by distance $d$, the magnetic force on a length $L$ is

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

where:

• $\mu_0 = 4\pi \times 10^{-7}\,\text{N/A}^2$
• $I_1 = 2\,\text{A}$
• $I_2 = 3\,\text{A}$
• $d = 5\,\text{cm} = 0.05\,\text{m}$

2. Which length should be used?

Wire $X$ has length $50\,\text{cm} = 0.5\,\text{m}$ and wire $Y$ has length $5\,\text{m}$.

Since wire $X$ is shorter, only the portion of wire $Y$ parallel to wire $X$ experiences the mutual force as given in the options. So we take

$L = 0.5\,\text{m}$

3. Substitute values

$F=\frac{(4\pi \times 10^{-7})(2)(3)(0.5)}{2\pi(0.05)}$

Simplify:

$F=\frac{(4\pi \times 10^{-7})(3)}{2\pi(0.05)}$

$F=\frac{12\pi \times 10^{-7}}{0.1\pi}$

$F=120 \times 10^{-7}$

$F=1.2 \times 10^{-5}\,\text{N}$

4. Direction of force

The currents are in the same direction, so the two parallel wires attract each other.

Hence, force on wire $Y$ is towards wire $X$.

5. Option check

• A: $1.2 \times 10^{-5}\,\text{N}$ towards $X$ ✅
• B: wrong magnitude and wrong direction ❌
• C: wrong magnitude ❌
• D: wrong magnitude ❌

Therefore, the correct option is A.