First, find the electric and magnetic fields generated by the individual wires. You have probably already done this if you are working on this problem, but if not, they are easy to find using a Gaussian surface and an Amperian loop:
$$ \oint \mathbf{E}\cdot d\mathbf{a} = \frac{Q_{enc}}{\epsilon_0} \quad\quad\quad\quad \oint \mathbf{B}\cdot d\mathbf{l} = \mu_0 I_{enc}$$
Once you have found the electric/magnetic fields generated by each wire individually, choose one wire (call it wire 1) and determine the force it imparts on the other (wire 2) using the Lorentz force law:
$$ \mathbf{F} = q(\mathbf{E}+\mathbf{v}\times\mathbf{B}) $$
(Where $\mathbf{v}$ is the velocity vector of the charge). Notice, however, that the force on wire 2 is not constant along the wire, so you'll have to look at each little segment $d\mathbf{l}$ of wire 2 separately. In each little segment there is a charge $\lambda dl$ (where $\lambda$ is the charge density on wire 2, and $dl$ is the magnitude of $d\mathbf{l}$). Thus, for this little piece of wire 2, the Lorentz force law becomes
$$ d\mathbf{F} = \lambda dl(\mathbf{E}+\mathbf{v}\times\mathbf{B}) $$
(Now $\mathbf{v}$ is the velocity vector of the charge in the little segment $d\mathbf{l}$ that you're looking at). If you look at the part $\lambda dl(\mathbf{v}\times \mathbf{B})$, you should recognize $\lambda \mathbf{v}$ as a current. Now we have
$$ d\mathbf{F} = dl(\lambda \mathbf{E}+\mathbf{I}\times\mathbf{B}) $$
(Recall that $\mathbf{I}$ is the current vector, which in your case is the same for every little segment $d\mathbf{l}$). Now we just have to think about what's going on. As you march down wire 2, each little segment that you pass (of length $dl$) experiences the force above (which you can calculate explicitly once you have found $\mathbf{E}$ and $\mathbf{B}$). Therefore to get the total force on wire 2 you would just integrate the above for all of wire 2. You should find that $\mathbf{E}$ and $\mathbf{B}$ depend on $s$, so you should think about the relationship between $dl$ and $ds$.
I hope this helps get you started.
I think there is a confusion between the concept of electromagnetic (EM) waves and EM field. Think about this two concept as the analogous of sound waves and air pressure. The first is the radiation, and the second is the medium where the radiation propagates.
Now, consider your system: a conducting wire containing a non-zero current. If the current is constant, there are no EM waves radiating from the wire, but there is only a stationary (constant in time) magnetic field, described by the Biot-Savart law (see Wikipedia).
It is important to understand that in the stationary case, electric and magnetic field are largely independent (if one sticks within a fixed reference frame), and their directions can be perpendicular, parallel, or anything in between. To see this, consider a region of space near a charged object and a magnet. The electrical field of the charge and the magnetic field of the magnet can assume different orientations in each point of the space, depending on the relative positions and orientations of the magnet and of the magnetic field.
EM waves are the excitations of the EM field propagating in space and time. They are typically the consequence of moving charges which are accelerating or changing directions. For example, if the current in the wire is alternating current (variable in time), conducting electrons do not have a constant speed[*] and produce EM radiation.
An EM wave is an oscillating solution of the Maxwell equations. These equations mandate that the magnetic and the electric components of this solutions are perpendicular.
However, since Maxwell equations are linear, any linear superposition of solutions of the Maxwell equations is still a solution. Therefore, one can consider, for example an EM wave superimposed to a stationary electric field. In this case the electric field and the magnetic field are not perpendicular.
[*] Ok, this is an oversimplification, because electrons don't have a constant field even in the case of stationary current (DC), due to scattering with the ion lattice. However this scattering processes are not sufficient to produce a macroscopically observable EM radiation.
Best Answer
“The paper” is probably just meaning the table. So let's put the wire vertically going in the up-down direction. Using the right-hand-rule we obtain that the magnetic field will be circles in the plane of the table/paper.
The compass will align itself to the magnetic field lines. Therefore the rotating compass should be positioned such that it points towards the wire or parallel to it. When you turn on the magnetic field, it can align itself with the magnetic field lines.
Take a look at some video of the experiment. There you can see that the magnet will align itself perpendicular to the wire and parallel to the plane that the wire is perpendicular to (that sentence is probably unnecessarily complex).