Non-circular motion of a particle in a perpendicular constant magnetic field

[feynman1]

There's a constant magnetic field B. If a particle is acted on by a force qv*B (* cross) only, and the initial velocity v0 is normal to B, is the motion certainly a circular one (for any m, q, v0)?
mv''=qv*B
If one solves this equation (vector), it doesn't seem obvious.

 

[A.T.]

There's a constant magnetic field B. If a particle is acted on by a force qv*B (* cross) only, and the initial velocity v0 is normal to B, is the motion certainly a circular one (for any m, q, v0)?
mv''=qv*B
If one solves this equation (vector), it doesn't seem obvious.

If acceleration has a constant magnitude and is always perpendicular to velocity you get uniform circular motion.

 

[vanhees71]

The correct equation (in non-relativistic approximation) is
$$m \dot{\vec{v}}=-\frac{q}{c} \vec{B} \times \vec{v}.$$
Since the cross product with constant $\vec{B}$ describes an infinitesimal rotation, it's immediately clear that the particle moves on a circle.

The detailed calculation is as follows: Choose the reference frame such that $\vec{B}=B \vec{e}_3$. Then the EoM reads in components
$$\dot{v}_1=\frac{q B}{mc} v_2=\omega v_2, \\
\dot{v}_2=-\frac{q B}{mc} B v_1=-\omega v_1, \\
\dot{v}_3=0.$$
This means that
$$v_3(t)=v_{30}=\text{const}.$$
To solve the other two equations, take the time derivative of the first equation and insert the 2nd equation:
$$\ddot{v}_1=-\omega^2 v_1$$
The general solution is
$$v_1(t)=C_1 \cos(\omega t) + C_2 \sin(\omega t).$$
From the 1st equation you get
$$v_2=\frac{1}{\omega} \dot{v}_1=-C_1 \sin(\omega t) + C_2 \cos(\omega t).$$
The initial condition in #1 can be used to orient the coordinate system such that $\vec{v}_0=v_0 \vec{e}_1$, i.e., $v_1(0)=v_0$, $v_2(0)=0$, $v_3(0)=v_{30}=0$. This gives
$$C_2=0, \quad C_1=v_0,$$
and thus
$$\vec{v}(t)=\begin{pmatrix} v_0 \cos(\omega t) \\ -v_0 \sin(\omega t) \\ 0 \end{pmatrix}.$$
Integrating once more
$$\vec{x}(t)=\vec{x}_0 + \frac{v_0}{\omega} \begin{pmatrix} \sin (\omega t) \\ \cos(\omega t) \\ 0 \end{pmatrix}.$$
This is indeed a circle (the particle running clockwise in the 12-plane if $q>0$ and thus $\omega>0$). The radius of the circle is
$$R=\frac{v_0}{\omega}=\frac{m c v_0}{q B}.$$

 

[feynman1]

The correct equation (in non-relativistic approximation) is
$$m \dot{\vec{v}}=-\frac{q}{c} \vec{B} \times \vec{v}.$$
Since the cross product with constant $\vec{B}$ describes an infinitesimal rotation, it's immediately clear that the particle moves on a circle.

The detailed calculation is as follows: Choose the reference frame such that $\vec{B}=B \vec{e}_3$. Then the EoM reads in components
$$\dot{v}_1=\frac{q B}{mc} v_2=\omega v_2, \\
\dot{v}_2=-\frac{q B}{mc} B v_1=-\omega v_1, \\
\dot{v}_3=0.$$
This means that
$$v_3(t)=v_{30}=\text{const}.$$
To solve the other two equations, take the time derivative of the first equation and insert the 2nd equation:
$$\ddot{v}_1=-\omega^2 v_1$$
The general solution is
$$v_1(t)=C_1 \cos(\omega t) + C_2 \sin(\omega t).$$
From the 1st equation you get
$$v_2=\frac{1}{\omega} \dot{v}_1=-C_1 \sin(\omega t) + C_2 \cos(\omega t).$$
The initial condition in #1 can be used to orient the coordinate system such that $\vec{v}_0=v_0 \vec{e}_1$, i.e., $v_1(0)=v_0$, $v_2(0)=0$, $v_3(0)=v_{30}=0$. This gives
$$C_2=0, \quad C_1=v_0,$$
and thus
$$\vec{v}(t)=\begin{pmatrix} v_0 \cos(\omega t) \\ -v_0 \sin(\omega t) \\ 0 \end{pmatrix}.$$
Integrating once more
$$\vec{x}(t)=\vec{x}_0 + \frac{v_0}{\omega} \begin{pmatrix} \sin (\omega t) \\ \cos(\omega t) \\ 0 \end{pmatrix}.$$
This is indeed a circle (the particle running clockwise in the 12-plane if $q>0$ and thus $\omega>0$). The radius of the circle is
$$R=\frac{v_0}{\omega}=\frac{m c v_0}{q B}.$$

Sorry I later realized the result was obvious but I was thinking of sth different.
BTW where does the c come from in your governing equation?

 

[Doc Al]

BTW where does the c come from in your governing equation?

The c appears when using Gaussian (cgs) units. Not when using SI units.

 

[vanhees71]

Sorry, I should always say, which units I use. Here I used Heaviside-Lorentz (or Gaussian, because there's no difference in the Lorentz-force formula in both CGS systems) units. It's simply that for it's easier to keep track of the correct dimensions in my equations, using the more natural CGS units than the SI, where I never know by heart, whether I should put some $\epsilon_0$ or $\mu_0$ somewhere ;-)).