A current-carrying wire joining two fixed points a and b is placed...

The magnetic force on a small element $d\vec{l}$ of the wire carrying current $I$ in a magnetic field $\vec{B}$ is given by $d\vec{F} = I(d\vec{l} \times \vec{B})$.

The total force on the wire from point $a$ to point $b$ is obtained by integrating the force over the entire length of the wire:

$\vec{F} = \int_a^{b} I(d\vec{l} \times \vec{B})$

Since the magnetic field $\vec{B}$ is uniform, it is constant and can be taken out of the integral:

$\vec{F} = I \left( \int_a^{b} d\vec{l} \right) \times \vec{B}$

The integral $\int_a^{b} d\vec{l}$ is simply the displacement vector $\vec{L}$ from the initial point $a$ to the final point $b$.

$\vec{F} = I(\vec{L} \times \vec{B})$

This expression shows that the net magnetic force depends only on the displacement vector between the two fixed points $a$ and $b$, the current $I$, and the uniform magnetic field $\vec{B}$. It does not depend on the actual path taken by the wire between these two points. Thus, the force is independent of the shape of the wire.