Incoming electrons

Since the magnetic field $B_0$ is in the z direction, every electron will enter a circular path with radius of $r= \frac {B_0e} {mv_F} = \frac {1 \times 1} {1\times 1} = 1$ . Let this path be $P_\theta$ where $\theta$ is the angle that an electron made with the x-axis when entering the plane.

The initial direction is tangential to the circular path and so is perpendicular to the radius of the path at the origin $O$ . Therefore, the center of path $P_\theta$ , let's call it $C_\theta$ is at a polar coordinate $(r,\theta)$ , in which $r$ is the distance from origin $O$ and $\theta$ is the angle measured from positive y-axis.

We now determine the position, $x_\theta$ , which this electron meets x-axis the second time (first time is when it is injected at origin $O$ ). Since electron goes in circular path, the center $C_\theta$ , origin $O$ , and $x_\theta$ , these 3 points make up a isosceles triangle. Thus, the distance $x_\theta$ from $O$ is $x_\theta=2r\sin\theta$ . Looking at the graph of function $\sin \theta$ for $0^\circ \leq \theta \leq 180 ^ \circ$ , these electrons will be most concentrated when $\frac {d} {d\theta} \sin \theta=0$ , which yields $\theta = 90 ^ \circ$ .

Substitute, we get $x_{max}=2r \sin 90 ^ \circ = 2r = 2.00$

We can think the 2-D graphene as a mass sheet on which electrons enter from one point (the origin) and can flow anywhere else except the lower half of the sheet. There are innumerable electrons flowing randomly on the sheet with the same speed $v_F$ . At one instance, a magnetic field of $B_0$ is turned on and suddenly all the electrons flow clockwise in a circle of $R=\frac{mv_F}{eB_0}$ .

Now let’s examine the conditions for the electrons to reach a point along the $x$ -axis. We draw many circles of radius $R$ that passes through the point. Any electrons that reaches that point along the $x$ -axis must be on one of those circles we draw and must have the velocity tangential to the circle, when the magnetic field is turned on. Since all the electrons on the lower half of the sheet are absorbed, there are only electrons flowing from upper half to lower half, but none from lower half to upper half. This means, all the electrons at any position along the $x$ -axis are flowing downward or along the axis, except for the origin, where electrons are injected, so they can have velocity either upward or downward or along the $x$ -aixs. (By downward, I mean velocity that has downward component, not limited to vertically downward.)

If we analyzing the circles more carefully, we can obtain the position along $x$ -axis that most electrons flow to by drawing a semi-circle, of which the left intersection with the axis is the origin and the right intersection is the position we want to find. Putting the origin at the left intersection of the semi-circle allows the electrons flowing vertically upward clockwise in the magnetic field to reach the position, which is $2R$ distance away from the origin.

Hence the distance from the origin is $2R=2\frac{mv_F}{eB_0}=2 m$

Every electron that's injected on the graphene sheet makes some random angle $c$ with the $x$ -axis, so it will hit the $x$ -axis at position $x=\frac{2mv_F}{eB}\sin(c)$ . If the electrons are injected with no preferential direction, the focusing point must be the place where the absolute value of $dc/dx$ is maximum, as $dx/dc=\frac{2mv_F}{eB}\cos(c)=0 \rightarrow c=\pi/2$ , so we get the focusing point is just $x_f=\frac{2mv_F}{eB}$ .

Use the numerical values and we get the result!