Generally, it isn't the case that the (net, total) magnetic flux threading the conducing loop is constant.
Note that the quote uses the word opposes which you have evidently taken to mean nullify. But in fact, if the solution requires an emf, then there must be a non-constant flux threading the loop.
When the conductive material forming the loop has non-zero resistivity, the solution must satisfy
$$\mathscr E = -\frac{\mathrm d\Phi}{\mathrm dt}$$
and
$$\mathscr E = R i$$
where $R$ is the resistance 'round the loop and $i$ is the current.
But
$$\Phi = \Phi_\textrm{ext} + \Phi_i = \Phi_\textrm{ext} + L i$$
where $L$ is the inductance of the conducting loop.
Combining the above yields
$$R i + L \frac{\mathrm di}{\mathrm dt} = -\frac{\mathrm d\Phi_\textrm{ext}}{\mathrm dt}$$
For example, consider the case that $\Phi_\textrm{ext} = \Phi_0$ is constant then the solution is
$$i(t) = I_0e^{-\frac{R}{L}t}$$
$$\Phi(t) = LI_0e^{-\frac{R}{L}t} + \Phi_0$$
$$\mathscr E(t) = -\frac{\mathrm d\Phi}{\mathrm dt} = -\left(-\frac{R}{L} \right)LI_0e^{-\frac{R}{L}t} =Ri(t)$$
So, this is all consistent and notice that the flux is not generally constant though the external flux is.
Now, consider the case that $\Phi_\textrm{ext}$ is increasing linearly with time
$$\Phi_\textrm{ext} = \Phi_0 + \phi t$$
then the particular solution is
$$i(t) = -\frac{\phi}{R}$$
$$\Phi(t) = \left(\Phi_0 -\frac{L}{R}\phi\right) + \phi t$$
$$\mathscr E = -\phi = Ri $$
Again, this is consistent and the flux is not generally constant. However, note that this result depends on $R$ being non-zero.
For the $R = 0$ case, we see that
$$L \frac{\mathrm di}{\mathrm dt} = -\frac{\mathrm d\Phi_\textrm{ext}}{\mathrm dt}$$
or
$$\Phi_i = - \Phi_\textrm{ext} + \mathrm{constant}$$
$$\mathscr E = -\frac{\mathrm d\Phi}{\mathrm dt} = 0$$
thus we conclude that, for a perfectly conducting loop, the magnetic flux threading the loop is constant.
Imagine the wire is infinitely long, at first. In that context it's easy to see that when the particle is far, far away from the loop, it will have practically no influence over it (i.e. the magnetic field induced by the movement of that charged particle, which is infinitely far away from the loop, wouldn't induce any current on it).
The same reasoning would work the other way around: when the particle is moving near the loop, the induced magnetic field will induce a current on the loop.
So whether the wire is infinite or not, the result is the same: as the particle approaches the loop, the influence over it increases; when it moves away from it, it decreases.
This leads to two cases:
- When the particle moves from $A$ to the center of the wire, the magnetic field is increasing. By the right-hand rule, on the surface of the loop the magnetic field will have a direction towards you (going out of the paper). As it is increasing, by Lenz's law a current opposed to this change has to appear: the current in the loop is clockwise.
- When the particle moves away from the loop (towards $B$), the magnetic field is decreasing. The direction is still the same (the particle keeps moving in the same direction). But now, by Lenz's law, the current on the loop must be counterclockwise.
Notice that the direction of the magnetic field is the same in both cases. It is its derivative that changes the sign, leading to the change in the direction of the current on the loop.
Best Answer
The book is wrong. The flux through the loop into the page is decreasing as the wire slides to the left. A clockwise current is induced in the loop, increasing the strength of the magnetic field into the page, and thus offsetting the change in flux.
The same answer is arrived at by considering the Lorentz force: charge carriers in the moving wire are subject to a force due to the fact that they're moving through a magnetic field. By $\vec{F}=q\vec{v} \times \vec{B}$, the positive charge carriers experience a force pushing them toward the bottom of the page, and the negative charge carriers experience a force pushing them toward the top of the page. Thus the direction of the flow of current is clockwise.