What does "point of application of force" mean in the given context? I faced a particular conceptual doubt while solving a textbook problem. I will initially write the complete question in my textbook and then try to boil it down to a single conceptual doubt so that it complies with the rules of Physics Stack Exchange.

Original Question:
A finite conductor $CD$ carrying current $i$ is placed near a fixed
  very long wire $AB$ current carrying $i_0$ as shown in the figure.
  Find the point of application and magnitude of the net ampere force on
  the conductor $CD$. What happens to the conductor $CD$ if it is free
  to move? (Neglect gravitational field)


My conceptual doubt is : What does "point of application of force" mean ? How to find it? 
According to me the force doesn't act on a single point but on the whole wire $CD$, the force being maximum at end $C$ and minimum at end $D$.

P.S: I hope the question complies with the rules of the site. If not, please inform me, so that I can try to improve/re-frame the question. Thank you. 
 A: Best guess, the point of application is an average position weighted by how much force is applied.
If the force was uniform, then that point would be the center.  Since it's not, it's stronger closer to the long wire, the point will be moved to the left of center a bit.  This means that not only is there a net force on the system, but also a net torque.
Try 
$$ x_{\rm center} = \frac{1}{F_{total}} \int  x\,\,{\rm d}F $$
A: Think about gravity for example. It acts on every point of an object, and one finds the center of gravity by doing a "weighted" sum of the point locations.
$$ x_{\rm CM} = \frac{ \sum x \Delta m}{\sum \Delta m} $$
If gravity varied by location as $g(x)$ then the above would be
$$ x_{\rm CM} = \frac{ \sum x \Delta m\, g(x)}{\sum \Delta m\, g(x)}= \frac{ \int x \,g(x)\,{\rm d}m}{\int g(x)\,{\rm d}m} $$
Now instead of gravity, substitute in the EM force $F(x)$
$$ x_{\rm CM} = \frac{ \int x \,F(x)\,{\rm d}m}{\int F(x)\,{\rm d}m} = \frac{ \int x \,F(x)\,{\rm d}x}{\int F(x)\,{\rm d}x} $$
NOTE: ${\rm d}m = \rho A {\rm d}x $ and the constant cross-section $A$ and density cancel each other between the top and the bottom of the fraction.
