Electron equations of motion through an uniform magnetic field

[libelec]

Homework Statement

An electron enters a zone of uniform magnetic field [tex]\vec B = 0,4T{\rm{ }}\hat j[/tex] with velocity [tex]{\vec V_0} = {10^5}m/s{\rm{ }}\hat i[/tex]. 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 = [tex]q\vec V \otimes \vec B[/tex]
- Newton's Second Law = [tex]\sum {\vec F} = m\frac{{{\partial ^2}\vec r}}{{\partial {t^2}}}[/tex]
- Conservation of Kinetic Energy = [tex]\Delta {E_k} = {W_{all{\rm{ }}forces}}[/tex]

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:

[tex]\vec F = q\vec V \otimes \vec B = m\frac{{d\vec V}}{{dt}}[/tex]
-[tex] 0 = m\frac{{d{V_x}}}{{dt}}[/tex]
-[tex] 0 = m\frac{{d{V_y}}}{{dt}}[/tex]
-[tex] q{V_x}B = m\frac{{d{V_z}}}{{dt}}[/tex]

Then

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

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:

[tex]\vec F = q\vec V \otimes \vec B = q{V_x}B{\rm{ }}\hat i[/tex]

I should rewrite the Lorentz Force this way?:

[tex]\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[/tex]

 

[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:

[tex]\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}
[/tex]

I calculate:

[tex]\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}
[/tex]

[tex]\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}
[/tex]

[tex]\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}
[/tex]

[tex]\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}
[/tex]

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

[tex]{V_x} = 50\cos \left( {\frac{{qB}}{m}t} \right)
[/tex]

Then I solve that:

[tex]\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}

[/tex]

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:

[tex]

\frac{dV_x}{dt} = -\frac{qV_zB}{m}
[/tex]

ehild

 

[libelec]

Thank you very much.