Answer
$2.5 \cdot 10^{-3}$ N rightward
Work Step by Step
We see that in the circuit, the magnetic field is going into the page on the rightmost side. We can calculate the magnetic force by:
$\displaystyle \vec{F}_B = I\vec{L} \times \vec{B}$
The force is only on the right side of the circuit (only the vertical part that is 10 cm) because the horizontal parts on the top and bottom of the circuit in which the magnetic field passes through have two forces components that act in opposite directions and therefore cancel each other out.
We can find the current of the system by using Kirchhoff's Loop Rule,
$\varepsilon - IR = 0$
$\displaystyle I = \frac{\varepsilon}{R} = \frac{15}{3} = 5$ A
$\therefore\displaystyle \vec{F}_B = (5)(0.01)(50\cdot10^{-3})\sin(90) = 2.5\cdot10^{-3}$ N
By the right-hand rule, the direction of the force is to the right since the magnetic field points into the page and the current points downward.
