The force on an electron is,(adsbygoogle = window.adsbygoogle || []).push({});

[tex]F\; =\; qV\times B[/tex]

where F is the force, V is velocity and B is the magnetic field. All are vectors except the charge of the electron q.

Expressing this as a cross product,

[tex]F\; =\; q\left[ \begin{array}{ccc} x^{\wedge } & y^{\wedge } & z^{\wedge } \\ V_{x} & V_{y} & V_{z} \\ B_{x}\left( t \right) & B_{y}\left( t \right) & B_{z}\left( t \right) \end{array} \right]\; =\; q\left( V_{y}B_{z}\left( t \right)\; -\; V_{z}By\left( t \right) \right)x^{\wedge }\; \; -\; q\left( V_{x}B_{z}\left( t \right)\; -\; V_{z}B_{x}\left( t \right) \right)y^{\wedge }\; +\; q\left( V_{x}B_{y}\left( t \right)\; -\; V_{y}B_{x}\left( t \right) \right)z^{\wedge }[/tex]

The x^ means a unit vector. Now we know that,

[tex]F\; =\; ma\; \; \; \; \; \; or\; \; \; \; \; \; a\; =\; \frac{F}{m}[/tex]

Lets define q/m = a

Now we have three differential equations,

[tex]\ddot{x}\; =\; a\left( \dot{y}B_{z}\left( t \right)\; -\; \dot{z}B_{y}\left( t \right) \right)[/tex]

[tex]\ddot{y}\; =\; a\left( \dot{x}B_{z}\left( t \right)\; -\; \dot{z}B_{x}\left( t \right) \right)[/tex]

[tex]\ddot{z}\; =\; a\left( \dot{x}B_{y}\left( t \right)\; -\; \dot{y}B_{x}\left( t \right) \right)[/tex]

There we have it. A system of three differential equations and I am stumped. Any takers?

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# Extremely difficult System of Differentil Equations

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