# Motion of a particle in a magnetic field.

• Oct 14th 2008, 02:50 PM
0-)
Motion of a particle in a magnetic field.
The only force acting on a particle with charge e and velocity v, in a constant magnetic field B, is ev × B. Also, r(0)=0 and $\dot{\textbf{r}}(0)=\textbf{V}$.

I've shown that: $m\dot{\textbf{r}}=e\textbf{r} \times \textbf{B} + m\textbf{V}$.

Now the question asks to find equations for the path of the particle if B=(B,0,0) , V=(v_1,v_2,0) and r=(x,y,z).

I evaluated the cross product and equated the components to get:

$\dot{x}=v_1$
$\dot{y}=\frac{eB}{m}z + v_2$
$\dot{z}=\frac{eB}{m}y$

By solving the second and third equations, I generate equations involving exponentials but the answer has equations involving sin and cos.

Can someone see where I've gone wrong? Thanks in advance.
• Oct 14th 2008, 08:06 PM
mr fantastic
Quote:

Originally Posted by 0-)
The only force acting on a particle with charge e and velocity v, in a constant magnetic field B, is ev × B. Also, r(0)=0 and $\dot{\textbf{r}}(0)=\textbf{V}$.

I've shown that: $m\dot{\textbf{r}}=e\textbf{r} \times \textbf{B} + m\textbf{V}$.

Now the question asks to find equations for the path of the particle if B=(B,0,0) , V=(v_1,v_2,0) and r=(x,y,z).

I evaluated the cross product and equated the components to get:

$\dot{x}=v_1$
$\dot{y}=\frac{eB}{m}z + v_2$
$\dot{z}=\frac{eB}{m}y$

By solving the second and third equations, I generate equations involving exponentials but the answer has equations involving sin and cos.

Can someone see where I've gone wrong? Thanks in advance.

$\dot{z} = {\color{red}-} \frac{eB}{m}y$.
• Oct 15th 2008, 05:25 AM
0-)
Quote:

Originally Posted by mr fantastic
$\dot{z} = {\color{red}-} \frac{eB}{m}y$.

Of course! I can't believe I went through it twice and made the same mistake both times.

Thank you for your help.