In the circuit depicted in the figure, a current of 5.0 A is...

Given:
Current, $I = 5.0 \text{ A}$
Length of wire PQ, $L = 50 \text{ cm} = 0.5 \text{ m}$
Magnetic field, $B = 0.20 \text{ T}$

The magnitude of the magnetic force on the current-carrying wire is given by:
$F = ILB \sin\theta$

Since the magnetic field is perpendicular to the plane of the circuit, $\theta = 90^\circ$.
$F = 5.0 \times 0.5 \times 0.20 = 0.50 \text{ N}$

In the given circuit diagram, the longer line of the battery symbol represents the positive terminal, so the current flows in a clockwise direction. Therefore, the current in wire PQ flows from P to Q (towards the right).

The magnetic field is directed into the plane of the paper. Using the right-hand rule for the cross product $\vec{F} = I(\vec{L} \times \vec{B})$ (or Fleming's Left-Hand Rule), with the current vector $\vec{L}$ pointing right and the magnetic field $\vec{B}$ pointing into the page, the resulting magnetic force is directed upwards.

An upward force on wire PQ points towards the inside of the closed circuit loop.

Thus, the magnetic force is $0.50 \text{ N}$ towards the inside of the circuit.