A rigid wire consists of a semicircular portion of radius R and two...

The magnetic force on a current-carrying wire in a uniform magnetic field is given by $\vec{F} = i(\vec{L}_{eff} \times \vec{B})$, where $\vec{L}_{eff}$ is the displacement vector from the initial point to the final point of the wire in the magnetic field.

The wire consists of two vertical straight sections and a semicircular section. The straight sections carry current in opposite directions (one upwards, one downwards). The magnetic forces on these straight sections are equal in magnitude and opposite in direction (pointing horizontally outwards), thus cancelling each other out.

For the semicircular portion, the effective length is the straight-line distance between its ends, which is the diameter $2R$. The direction of $\vec{L}_{eff}$ is horizontally from left to right.

The magnetic field $\vec{B}$ is uniform and directed into the plane of the paper.

The magnitude of the net force is $F = i|\vec{L}_{eff}||\vec{B}|\sin(90^{\circ}) = i(2R)B = 2iRB$.

Using the right-hand rule for the cross product $\vec{L}_{eff} \times \vec{B}$, the direction of the force is upwards in the plane of the figure.

Therefore, the net magnetic force on the wire is $2iRB$, upward.