Torque is given by $$\vec{\tau}=\vec{r}\times\vec{F}$$ and it is a vector quantity. This means the torque has to be perpendicular to both the force and the radius. However, this doesn't seem to be the case for the question below:
Where the answer says the torque goes into the page
Doesn't this contradict the definition of torque given above?
While the torque equation and answer is correct, the statement you gave is wrong, I.e., torque is not necessarily perpendicular to the radius.
It is perpendicular to the displacement vector $r$, between point of application and centre. Here force on wire AD is downwards and BC upwards. So it constitutes a couple force, where there is net force but there is net torque.
The wire rotates clockwise as seen from front. You may want to refer to how a electric motor works and the coil rotates.
Hope it helps.
The question is flawed in that the best you can do is to find the magnitude $(1.68\rm\,N)$ and direction (into the screen) of the force on side $BC$.
What is not given in the question is the axis about which you need to find the torque.
Let us assume it is the vertical axis which bisects the coil about which the torque is to be found.
Then your $\vec r$ has a magnitude of $0.03\,\rm m$ and is directed to the right.
So the direction of the torque is
$\tau = \vec r \times \vec F \Rightarrow \hat {\rm right} \times \hat{\text {into screen}} = \hat {up}$ ie anticlockwise when look up from the bottom.
Of the torque definition can answer the question.
Each infinitesimal segment of wire with length ${\rm d}s$, direction $\vec{e}$ and located at $\vec{r}$ has an infinitesimal force ${\rm d}\vec{F}$ applied to it.
The net force applied is
$$\vec{F}_{\rm net} = \int {\rm d} \vec{F}$$
The net torque on the wire is adding up all the little segments.
$$\vec{\tau}_{\rm net} = \int \vec{r} \times \,{\rm d}\vec{F}$$
and due to magnetism the force is defined for each segment as ${\rm d}\vec{F} = (\vec{i} \times \vec{B}) {\rm d}s$
Now to answer your question, you need to evaluate the integral
$$ \vec{\tau}_{\rm net} = \int_0^\ell \vec{r} \times (\vec{i} \times \vec{B}) {\rm d}s$$
where the location varies along the wire $\vec{r} = \vec{r}(s)$.
Note that if the end-points of the wire are designated at $\vec{r}_1$ and $\vec{r}_2$ and the wire is a straight line you have
$$ \vec{r}(s) = \left(1 - \tfrac{s}{\ell}\right) \vec{r}_1 + \left(\tfrac{s}{\ell}\right) \vec{r}_2 $$
with the net torque
$$ \vec{\tau}_{\rm net} = \left( \int_0^\ell \vec{r} {\rm d}s \right) \times (\vec{i} \times \vec{B}) = \frac{ \vec{r}_1 + \vec{r}_2}{2} \times (\vec{i} \times \vec{B})$$
which is interpreted as the force acting on the center of mass of the wire, and then using $$\tau_{\rm net} = \vec{r}_{\rm com} \times \vec{F}_{\rm net}$$
In a rotating object, the velocity and acceleration vectors are different for each point and are continuously changing direction. To deal with this, the rotational quanties are defined as vectors directed along the axis of rotation which is usually not changing direction. This includes the torque vector (which you normally want to be in the same direction as the angular acceleration vector). The direction of the torque vector as indicated by a cross product is consistent with this convention. In this problem, the force on the right side of the loop is into the page, but the torque (relative to the axis of rotation) as a vector is up along the axis.
