1. **Problem statement:** A positively charged particle moves horizontally with velocity $\mathbf{v}$ along a straight long wire carrying current $I$. We need to find the direction of the magnetic force exerted on the particle. 2. **Relevant formula:** The magnetic force $\mathbf{F}$ on a charged particle moving in a magnetic field $\mathbf{B}$ is given by the Lorentz force law: $$\mathbf{F} = q \mathbf{v} \times \mathbf{B}$$ where $q$ is the charge, $\mathbf{v}$ is the velocity vector, and $\mathbf{B}$ is the magnetic field vector. 3. **Magnetic field due to a current-carrying wire:** The magnetic field around a long straight wire carrying current $I$ is given by the right-hand rule and has magnitude: $$B = \frac{\mu_0 I}{2 \pi r}$$ where $r$ is the distance from the wire and $\mu_0$ is the permeability of free space. 4. **Direction of $\mathbf{B}$:** Using the right-hand rule, if current $I$ flows along the wire, the magnetic field circles the wire. At the location of the particle moving parallel to the wire, $\mathbf{B}$ is perpendicular to $\mathbf{v}$. 5. **Direction of force $\mathbf{F}$:** Since $\mathbf{F} = q \mathbf{v} \times \mathbf{B}$ and $q$ is positive, the force direction is given by the right-hand rule for the cross product of $\mathbf{v}$ and $\mathbf{B}$. 6. **Conclusion:** The magnetic force is perpendicular to both the velocity and the magnetic field directions. For a particle moving parallel to the wire, the force is directed either up or down depending on the orientation of $I$ and $\mathbf{v}$. Assuming the current $I$ flows to the right and the particle moves to the right above the wire, the magnetic field $\mathbf{B}$ points into the page (from the observer). Then $\mathbf{v} \times \mathbf{B}$ points upward. **Final answer:** The magnetic force exerted on the particle is **Up**.