# Electron in magnetic field and center of circumference

1. Feb 9, 2013

### bznm

1. The problem statement, all variables and given/known data

We have an electron in 3D space in which there is also a magnetic uniform field B=Bz.
At t=0, we have z=0 and $v_z=0$.
We have to find the center of the circumference travelled by the electron using only x, y, v_x and v_y (coordinates and velocity of the electron at the general istant t) .

2. Relevant equations
$$L_F= qv×B$$ : this is the equation of Lorentz force.

3. The attempt at a solution
The first thing that I have done was finding the radius: $R=\frac{mv}{qB}$
Then, I have thought that I could obtain the center considering:

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

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

.. but probably this way is wrong.

And so I have thought to use Lorentz equation

$$F_{L,x}=q v_y B$$ $$F_{L,y}=-q v_x B$$ $$F_{L,z}=0$$
but I don't know how I can obtain the center using this equation... what can I do?

Thank you so much!!

2. Feb 9, 2013

### Staff: Mentor

If the electron moves in some direction at t=0, where is the center of the circle? In the direction of motion, behind it, left*, right*, something in between?
*as seen from "above" (positive z), for example

How can you find a point in that direction with a distance of R?

3. Feb 9, 2013

### bznm

on the left? (negative charge move in anticlockwise..)

I have thought to use the euclidean distance between points, but I obtain only one relation and I don't obtain the coordinates of the center using the velocities.. can you help me? thanks a lot!

4. Feb 9, 2013

### Staff: Mentor

Hmm, let's view that from below, along the magnetic field. Ok, to the left, so at a 90°-angle to the direction of motion. Do you know how you can find an orthogonal line in 2 dimensions? The velocity components will help. And the euclidian distance will be used, too.

5. Feb 9, 2013

### bznm

If I consider the line that have the direction of motion, I obtain
$$y-y_T=y'(x-x_T)$$

and the radius belongs to the line

$$y-y_T=-1/y'(x-x_T)$$

(y' is v_y, isn't it?)

and the euclidean distance between the position of the electron at generical istant t (x_T,y_T) and the center of the circumference (x_c,y_c) is

$$sqrt ((x_T-x_c)^2+(y_t-y_c)^2))=R$$

now I don't know what I have to do...

6. Feb 9, 2013

### Staff: Mentor

What is yT?

For simplicity, let's use x=y=0. You can add them afterwards.

If the velocity vector is (x',y'), can you find an orthogonal vector?

7. Feb 9, 2013

### bznm

If the vector a=(a1,a2) si orthogonal to (x',y') then
a1*x'+a2*y'=0..

8. Feb 9, 2013

### Staff: Mentor

You can use this to generate some orthogonal vector. In 2 dimensions, there is a very nice way to chose a1 and a2.