1. **Problem statement:** We have three currents $I_1$, $I_2$, and $I_3$ in a uniform magnetic field $\vec{B}$ pointing horizontally to the right. We want to find which current experiences the maximum magnetic force and which experiences zero force. 2. **Recall the magnetic force formula:** The magnetic force on a current-carrying wire is given by $$\vec{F} = I \vec{L} \times \vec{B}$$ where $I$ is the current, $\vec{L}$ is the length vector of the wire in the direction of current, and $\vec{B}$ is the magnetic field. 3. **Magnitude of force:** The magnitude of the force is $$F = I L B \sin\theta$$ where $\theta$ is the angle between $\vec{L}$ and $\vec{B}$. 4. **Analyze each current direction relative to $\vec{B}$:** - $I_1$: arrow tilted down left, so $\theta$ is approximately $135^\circ$ (since $\vec{B}$ points right, and $I_1$ points down-left). - $I_2$: arrow straight downward, so $\theta = 90^\circ$. - $I_3$: arrow horizontal right, so $\theta = 0^\circ$ (same direction as $\vec{B}$). 5. **Calculate $\sin\theta$ for each:** - $\sin 135^\circ = \sin 45^\circ = \frac{\sqrt{2}}{2} \approx 0.707$ - $\sin 90^\circ = 1$ - $\sin 0^\circ = 0$ 6. **Determine force magnitudes:** - $F_1 = I_1 L B \times 0.707$ - $F_2 = I_2 L B \times 1$ - $F_3 = I_3 L B \times 0$ 7. **Conclusion:** - Maximum force is on $I_2$ (since $\sin 90^\circ = 1$ is maximum). - Zero force is on $I_3$ (since $\sin 0^\circ = 0$). **Answer:** Maximum force on $I_2$, zero force on $I_3$. The given choice (Ⓐ) states maximum force on $I_1$ and zero force on $I_3$, which is incorrect based on the analysis. Hence, the current experiencing zero force is $I_3$, but the maximum force is on $I_2$, not $I_1$.