Electron equations of motion through an uniform magnetic field

[libelec]

Homework Statement

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?

Homework 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}}

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.

 

[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

 

[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}}<br /> {\hat i} &amp; {\hat j} &amp; {\hat k} \\<br /> {{V_x}} &amp; {{V_y}} &amp; {{V_z}} \\<br /> 0 &amp; B &amp; 0 \\<br /> \end{array}} \right| = - q{V_z}B{\rm{ }}\hat i + q{V_x}B{\rm{ }}\hat k

 

[ehild]

It is correct. Now you can set up the differential equations for V_x, V_y and V_z.

ehild

 

[libelec]

Could somebody correct me if I'm wrong?

Given the new set of differential equations:

\begin{array}{l}<br /> \frac{{d{V_x}}}{{dt}} = \frac{{q{V_z}B}}{m} \\ <br /> \frac{{d{V_y}}}{{dt}} = 0 \\ <br /> \frac{{d{V_z}}}{{dt}} = \frac{{q{V_x}B}}{m} \\ <br /> \end{array}<br />

I calculate:

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

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

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

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

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

{V_x} = 50\cos \left( {\frac{{qB}}{m}t} \right)<br />

Then I solve that:

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

I will get a similar expression for r_x, r_y and r_z, which translates (I think) into the equation of a circle.

Is this OK?

 

[ehild]

Just a small mistake:

<br /> <br /> \frac{dV_x}{dt} = -\frac{qV_zB}{m} <br />

ehild

 

[libelec]

Thank you very much.