# Electron equations of motion through an uniform magnetic field

1. Oct 31, 2009

### libelec

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

An electron enters a zone of uniform magnetic field $$\vec B = 0,4T{\rm{ }}\hat j$$ with velocity $${\vec V_0} = {10^5}m/s{\rm{ }}\hat i$$. Find the differential equations that govern its motion through the field, and solve them to find the equations of motion. What happens to its kinetic energy?
2. Relevant equations

- Lorentz Force = $$q\vec V \otimes \vec B$$
- Newton's Second Law = $$\sum {\vec F} = m\frac{{{\partial ^2}\vec r}}{{\partial {t^2}}}$$
- Conservation of Kinetic Energy = $$\Delta {E_k} = {W_{all{\rm{ }}forces}}$$

3. The attempt at a solution

I know that the answer should be that the electron's trajectory is a circle. But I can't get there throught the differential equations:

If I don't take the electron's weight into account, I have that the only force acting upon it is the Lorentz Force. Using Newton's Second Law:

$$\vec F = q\vec V \otimes \vec B = m\frac{{d\vec V}}{{dt}}$$
-$$0 = m\frac{{d{V_x}}}{{dt}}$$
-$$0 = m\frac{{d{V_y}}}{{dt}}$$
-$$q{V_x}B = m\frac{{d{V_z}}}{{dt}}$$

Then

-$${V_x} = {10^5}m/s$$
-$${V_y} = 0$$
-$$\frac{{q{V_x}B}}{m}t = {V_z}$$

I know there's something wrong: since the only force acting upon the electron is the Lorentz Force, being a central force (perpendicular to the trajectory), it doesn't do any work, the kinetic energy conserves and therefore the module of V should be constant. Which doesn't happen if the solution I found is true (I know it's wrong).

What's wrong with my resolution?

Thanks.

2. Nov 1, 2009

### ehild

You can not assume that the x component of the velocity stays constant. Decompose the Lorentz force into all components, assuming non-zero components of velocity, and discuss what you got.

ehild

3. Nov 1, 2009

### libelec

You say that instead of keeping the Lorentz Force this way:

$$\vec F = q\vec V \otimes \vec B = q{V_x}B{\rm{ }}\hat i$$

I should rewrite the Lorentz Force this way?:

$$\vec F = q\vec V \otimes \vec B = q\left| {\begin{array}{*{20}{c}} {\hat i} & {\hat j} & {\hat k} \\ {{V_x}} & {{V_y}} & {{V_z}} \\ 0 & B & 0 \\ \end{array}} \right| = - q{V_z}B{\rm{ }}\hat i + q{V_x}B{\rm{ }}\hat k$$

4. Nov 1, 2009

### ehild

It is correct. Now you can set up the differential equations for Vx, Vy and Vz.

ehild

5. Nov 1, 2009

### libelec

Could somebody correct me if I'm wrong?

Given the new set of differential equations:

$$\begin{array}{l} \frac{{d{V_x}}}{{dt}} = \frac{{q{V_z}B}}{m} \\ \frac{{d{V_y}}}{{dt}} = 0 \\ \frac{{d{V_z}}}{{dt}} = \frac{{q{V_x}B}}{m} \\ \end{array}$$

I calculate:

$$\begin{array}{l} {V_z} = \left( {\frac{{d{V_x}}}{{dt}}} \right)\frac{m}{{qB}} \\ {V_y} = 0 \\ \frac{{d{V_z}}}{{dt}} = \frac{{q{V_x}B}}{m} \\ \end{array}$$

$$\begin{array}{l} {V_z} = \left( {\frac{{d{V_x}}}{{dt}}} \right)\frac{m}{{qB}} \\ {V_y} = 0 \\ \frac{{d\left( {\frac{{d{V_x}}}{{dt}}\frac{m}{{qB}}} \right)}}{{dt}} = \frac{{q{V_x}B}}{m} \\ \end{array}$$

$$\begin{array}{l} {V_z} = \left( {\frac{{d{V_x}}}{{dt}}} \right)\frac{m}{{qB}} \\ {V_y} = 0 \\ \frac{{{d^2}{V_x}}}{{d{t^2}}}\frac{m}{{qB}} = \frac{{q{V_x}B}}{m} \\ \end{array}$$

$$\begin{array}{l} {V_z} = \left( {\frac{{d{V_x}}}{{dt}}} \right)\frac{m}{{qB}} \\ {V_y} = 0 \\ \frac{{{d^2}{V_x}}}{{d{t^2}}} = \frac{{{q^2}{B^2}}}{{{m^2}}}{V_x} \\ \end{array}$$

I propose the solution for Vx (should I include the 50 that way?):

$${V_x} = 50\cos \left( {\frac{{qB}}{m}t} \right)$$

Then I solve that:

$$\begin{array}{l} {V_x} = 50\cos \left( {\frac{{qB}}{m}t} \right) \\ {V_y} = 0 \\ {V_z} = - 50\sin \left( {\frac{{qB}}{m}t} \right) \\ \end{array}$$

I will get a similar expression for rx, ry and rz, which translates (I think) into the equation of a circle.

Is this OK?

6. Nov 1, 2009

### ehild

Just a small mistake:

$$\frac{dV_x}{dt} = -\frac{qV_zB}{m}$$

ehild

7. Nov 1, 2009

### libelec

Thank you very much.