A wire carrying a current i is positioned in the x – y plane along...

The magnetic force on a current-carrying wire in a uniform magnetic field is given by $\vec{F} = \int i (d\vec{l} \times \vec{B})$.

Since the magnetic field $\vec{B} = B \hat{k}$ is uniform, the total force depends only on the displacement vector $\vec{L}$ between the initial and final points of the wire.

The initial point at $x = 0$ is $(0, 0, 0)$.

The final point at $x = \lambda$ is $(\lambda, A \sin(2\pi), 0) = (\lambda, 0, 0)$.

The displacement vector is $\vec{L} = \lambda \hat{i}$.

The magnetic force is $\vec{F} = i (\vec{L} \times \vec{B}) = i (\lambda \hat{i} \times B \hat{k}) = -i \lambda B \hat{j}$.

The magnitude of the magnetic force is $|\vec{F}| = i \lambda B$.