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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:


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?

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Why don't you just define that the x=0 and y=0 are the origin of R? –  anna v Feb 8 '13 at 19:15
@annav It could be a good idea, but let's suppose that I can't do it.. ;) –  sunrise Feb 8 '13 at 20:05
Then you are solving a simple geometrical question, "given a point at x_c and y_c on the periphery of the circle with radius R where is the center in cartesian coordianates?". No need of differential equations. The charged part will advise you which of the two solutions to pick. –  anna v Feb 9 '13 at 4:39
have a look at analyzemath.com/CircleEq/Tutorials.html , the general equation in the x, y plane of a circle with known radius –  anna v Feb 9 '13 at 16:46
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2 Answers

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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).

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thanks for your help! :) which solutions do I have to transform into cartesian? –  sunrise Feb 9 '13 at 10:02
I'm really sorry, John, but I haven't any initial condition about $v_x$ and $v_y$ and I can't suppose anything.. How can I obtain $x_c$ and $y_c$ using only $x$,$y$ (coordinates of the electron at the general istant t) and $\dot x$, $\dot y$? I hope you'll have a little of patience again... –  sunrise Feb 9 '13 at 15:56
John, the @tag mechanism does not work in post (as opposed to comments). In addition, the asker always gets notified of new answers (and I think of new versions of existing answers). Finally we seek here not just to answer the asker but to provide a reference for future visitors. –  dmckee Feb 9 '13 at 16:31
@dmckee Thanks for your comments Re tag. I am new to this interesting site, still learning about it. Also thanks for the info with regards to the scope of the site. I shall restrain myself to simply giving hints, I really thought this is what I was doing. Regards. –  JKL Feb 9 '13 at 20:44
Thanks a lot for your patience and for your edits!! Your answer is perfectly clear! :) (just a note: when you edit an answer, the OP doesn't recive a notification.. if you want that the OP sees the edit, you have to tag him in a comment like "@OP updated answer") –  sunrise Feb 13 '13 at 22:07
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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}$.

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?

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Just after I finished my answer, I saw yours and we are giving, braodly speaking, the same advice. I have given you +1 for this coincidence! –  JKL Feb 8 '13 at 23:02
Thanks for your answer. I know that $F_{L_x}=qv_yB$, $F_{L,y}=-qv_x B$, $F_{L,z}=0$ but I don't know how I can 'link' these relations to the center of the circumference.. thanks again! –  sunrise Feb 9 '13 at 9:35
@sunrise, great! You're on the right track. See my edits –  zhermes Feb 9 '13 at 16:12
Thanks. I'm 'seeing' the problem clearer. I have obtained that $v_x=\alpha cos( \frac{qB}{m}t)+\beta (sin \frac{qB}{m}t)$. I hope that I haven't made mistakes. But..now? If I had $v_x$, I'd find $v_y$ and then integer and obtain $x$ and $y$.. –  sunrise Feb 10 '13 at 15:32
I can't understand how I can obtain something like $v_x=Acos \omega t$ starting from the differential equation $\frac{d v_x}{dt}=-v_x \omega ^2$, calling with $\omega$ $\frac {qB}{m}$. If I'm not wrong, the solution of this DE, is given by $v_x=\alpha cos \omega t+ \beta sin \omega t$. (I've just begun to study differential equations..) –  sunrise Feb 11 '13 at 12:04
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