A current of 5.0 A flows in a semicircular wire having a radius of...

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

For a semicircular wire of radius $R$, the magnitude of the displacement vector is the diameter of the semicircle, so $L = 2R$.

Since the magnetic field is perpendicular to the plane of the wire, the angle between $\vec{L}$ and $\vec{B}$ is $90^{{}^{{}^{\circ}}}$.

The magnitude of the magnetic force is $F = I L B \sin(90^{{}^{{}^{\circ}}}) = I(2R)B$.

Given $I = 5.0 \text{ A}$, $R = 5.0 \text{ cm} = 0.05 \text{ m}$, and $B = 0.50 \text{ T}$.

Substituting the values:

$F = 5.0 \times (2 \times 0.05) \times 0.50$

$F = 5.0 \times 0.10 \times 0.50 = 0.25 \text{ N}$