Electron in a uniform magnetic field and center of the circumference

Question

I have an electron in a 3D space and there is a uniform constant magnetic field $B$, oriented according to z-axes.

At $t=0$, we have $z=0$ and $\dot z =0$.

I have proved that the trajectory is a circumference and that the motion is uniform. Now I have to find the center of the circumference.

I have found the radius:

$$R=\frac{mv}{qB}$$

And I have thought that I could find the coordinates of the center considering:

$$R_x= \frac{m \dot x}{qB} \rightarrow x-x_c=\frac{m \dot x}{qB} $$

$$R_y= \frac{m \dot y}{qB} \rightarrow y-y_c=\frac{m \dot y}{qB} $$

Is it correct?

Answer 1

From your given information I realise that your problem is pure geometry than physics. You know the radius of the circle

$R={\frac {mv}{qB} }$

You need to know the initial position, $P_0(x_0, y_0)$ of the electron, and the direction of the velocity. Let as say these are given by

${\bf r}_0=x_0{\bf i}+y_0{\bf j}$

${\bf u}=u_x{\bf i}+u_y{\bf j}$

For simplicity you can take u to be ${\bf u}=v{\bf i}$ along the x-axis. Also let us assume the centre of the circle is at $(x_c, y_c)$. Then you need some vector algebra:

$({\bf r}_0-{\bf r}_c).{\bf u}=0$ because the radius and tangent are perpendicular.

$(x_0-x_c)^2+(y_0-y_c)^2=R^2$ general equation of a circle

Solve these two simultaneous equations to find $x_c$ and $y_c$. If you take simple initial conditions $P_0(0,0)$ and {${\bf u}=(v,0)$} then according to Fleming’s rule the centre of the circle must be at the point (0, R).

Answer 2

That is not correct. There is no reason for the x position to be linearly proportional to the x-velocity. In fact, because you know the motion is circular, you also know that when the x-position is at a maximum, the x-velocity is zero.

Because you have the correct expression for the orbital radius, and you know that the angular velocity is constant (assuming the magnetic field is constant), you can easily express the answer in polar coordinates, $\vec{R} = \left(R,\theta(t) \right)$.

In general, however, to find the position vector you would need to solve the differential equations of motion, where the force is provided by the lorentz force, i.e.

$ F_L = m\frac{dv}{dt} = q\vec{v} \times \vec{B}$.

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Edit:
As the OP posted in the comments, $F_x = q v_y B$ and $F_y = -q v_x B$, i.e.

$F_x: \frac{dv_x}{dt} = v_y \frac{qB}{m}$

$F_y: \frac{dv_y}{dt} = -v_x \frac{qB}{m}$

The next step is to try to find general expressions for $v_x$ and $v_y$. Again, it's best to use your knowledge of the general behavior of the solution (i.e. circular motion) to make a guess: try, something like $v_x = A\cos{\omega t}$. Using that, what do you get for $v_y$? Once you have both velocities, how can you find the $x,y$ positions?