Force on Current-Carrying Conductors question from JEE Main 2002 (AIEEE)

Answer: (c)

1. Field of wire 1 at the element: $B = \frac{\mu_0 i_1}{2\pi r}$, perpendicular to the plane containing both wires.
2. $d\vec F = i_2\,d\vec l \times \vec B$. Since $\vec B \perp d\vec l$, $dF = \frac{\mu_0 i_1 i_2\,dl}{2\pi r}$.
3. The component of this force along the perpendicular from wire 1 (the attracting part) is $\frac{\mu_0 i_1 i_2\,dl}{2\pi r}\cos\theta$, which is the form given in the key.
4. So the answer is (c) $\frac{\mu_0}{2\pi r} i_1 i_2\,dl\cos\theta$.