To solve this problem, we will use Kirchhoff's loop rule.$\\$Given circuit:$\\$• Resistors: $2 \, \Omega, 1 \, \Omega, 7 \, \Omega \\$• Batteries: $10 \, V$ and $5 \, V \\$Let's denote:$\\$• $I$ as the current flowing in the circuit.$\\$Apply Kirchhoff's loop rule:$\\$The sum of the potential differences around a closed loop is zero.$\\$Starting from point $A$ and moving clockwise:$\\ -10 \, V + 2I + 1I + 5 \, V + 7I = 0 \\$Simplify the equation:$\\ -5 + 10I = 0 \\$Solve for $I: \\ 10I = 5 \\ I = \frac{5}{10} \\ I = 0.5 \,$A$\\$The direction of the current is from $A$ to $B$ through $E. \\$Therefore, the magnitude and direction of the current is $0.5 \,$A from $A$ to $B$ through $E. \\$This corresponds to Option 2.