To solve this problem, we need to analyze the forces acting on the rod and use the concept of magnetic force.$\\$Given:$\\$• Mass per unit length of the rod $\lambda = 0.5 \,$kg/m$\\$• Angle of inclination $\theta = 30^\circ \\$• Magnetic field $B = 0.25 \,$T$\\$The rod is stationary, so the net force along the incline must be zero.$\\$The gravitational force component along the incline is given by:$\\ F_{gravity} = \lambda \cdot g \cdot \sin \theta \\$The magnetic force acting on the rod is given by:$\\ F_{magnetic} = I \cdot L \cdot B \cdot \sin \phi \\$Since the magnetic field is vertical, $\phi = 90^\circ,$ so $\sin \phi = 1. \\$For equilibrium, $F_{gravity} = F_{magnetic}. \\$Substitute the expressions:$\\ \lambda \cdot g \cdot \sin \theta = I \cdot L \cdot B \\$Since $L$ is the length of the rod, it cancels out from both sides.$\\$Rearrange to solve for $I: \\ I = \frac{\lambda \cdot g \cdot \sin \theta}{B} \\$Substitute the known values:$\\ I = \frac{0.5 \cdot 9.8 \cdot \sin 30^\circ}{0.25} \\$Calculate $\sin 30^\circ: \\ \sin 30^\circ = 0.5 \\$Substitute and simplify:$\\ I = \frac{0.5 \cdot 9.8 \cdot 0.5}{0.25} \\ I = \frac{2.45}{0.25} \\ I = 9.8 \,$A$\\$Therefore, the current flowing in the rod to keep it stationary is $9.8 \,$A$. \\$This corresponds to Option 4.