When an electron moves in uniform external magnetic field, with velocity not perpendicular, I understand that the magnitude of force is only due to the perpendicular component of velocity and is perpendicular to both components of velocity. But this force, though only due to the perpendicular component, acts on the particle. So why is it that the parallel component of velocity is not changed whereas the perpendicular component of velocity keeps changing its direction.

Is there a way to prove that its path is helical, using only vectors, and to show that it does not affect the parallel component? I've tried looking everywhere but haven't found any proof using vectors.


4 Answers 4


Let's assume the magnetic field vectors point in z-direction (or: let's call the direction the magnetic field vector points "z"). Then we have for the magnetic field:

$$\vec{B} = \begin{pmatrix}0\\0\\B\end{pmatrix}$$

and for the speed of the electron:

$$\vec{v} = \begin{pmatrix}v_x\\v_y\\v_z\end{pmatrix}$$

The Lorentz-force $\vec{F}$ due to a magnetic field is given by

$$\vec{F} = q\vec{v}\times\vec{B}$$ $$= q\begin{pmatrix}v_x\\v_y\\v_z\end{pmatrix}\times\begin{pmatrix}0\\0\\B\end{pmatrix}$$ $$= q\begin{pmatrix}v_y B - 0\\0 - v_x B\\0 - 0\end{pmatrix}$$ $$= qB\begin{pmatrix}v_y \\ - v_x \\0 \end{pmatrix}$$

So as you can see the force in z-direction is 0 and due to Newtons law ($F = m a$) the acceleration is as well, resulting in constant speed in z-direction.

If you now solve the differential equations

$$\frac{d}{dt}\begin{pmatrix}v_x \\ v_y \\v_z \end{pmatrix} = qB\begin{pmatrix}v_y \\ - v_x \\0 \end{pmatrix}$$

you will find that the path is indeed helical, i.e. that $$\vec{v}(t) = \begin{pmatrix}v_0 \sin(\omega t + \phi) \\ v_0 \cos(\omega t +\phi) \\v_z \end{pmatrix} $$ is a valid solution. (Disclaimer: might not be the only possible solution) $\omega$ here depends on q and B, while $\phi$ should depend on your coordinate-system and $\vec{v}(0)$.

For the differential equation you could do this:

$$\frac{d}{dt} v_x = qBv_y$$ $$\frac{d^2}{dt^2} v_x = qB\frac{d}{dt}v_y = -q^2B^2v_x$$ $$\frac{d^2}{dt^2} v_x = -q^2B^2v_x$$

So we are searching for a function whichs second derivative is the function itself but with a negative factor. So you either look it up in a table, do some fancy math-magic, assume it's an exponential function or just know from previous studies that its a linear combination of $\sin$ and $\cos$. Linear combinations of sin and cos can be represented as a phase $\phi$, so pick one randomly, add a phase, calculate $v_y = \frac{d}{dt} \frac{v_x}{qB}$ and you are done.

For a mathematical proof that this is the only solution or additional solutions read the pack insert and ask your doctor or mathematician. ;-)

  • $\begingroup$ What if there is velocity only along the z direction and along x direction ? I.e velocity along y direction is zero ? In that case there is force only along y direction. $\endgroup$
    – TESLA____
    Feb 7, 2016 at 11:56
  • $\begingroup$ At $t=0$, yes. But since you have a force in y-direction you will then have a velocity in y-direction as well. In the xy-plane this is exactly what causes a circular motion: a force always perpendicular to the motion. If you set $\phi=\frac{\pi}{2}$ in the given solution you end up with $v_x(0)=v_0, v_y(0)=0$ which describes this case. $\endgroup$
    – Anedar
    Feb 7, 2016 at 12:04
  • $\begingroup$ Ok. I get it. But could you show how to solve the differential equation ? $\endgroup$
    – TESLA____
    Feb 7, 2016 at 12:07
  • $\begingroup$ I edited in at least a rough solution. It should give you a hint how to find the given solution. $\endgroup$
    – Anedar
    Feb 7, 2016 at 12:23
  • $\begingroup$ Anedar, see my comment in the form of an answer (because to long for a comment). $\endgroup$ Feb 7, 2016 at 15:20

You can put the vectors in at the end because this an example of motions in different directions being independent of one another.

There is only one force acting on the charged particle and that if due to the component of velocity of the particle which is at right angles to the B-field.
Since that force is perpendicular to the direction of motion and of constant magnitude it provides a constant centripetal acceleration and the particle undergoes circular motion.

The component of velocity which is parallel to the B-field stays constant as there is no force in this direction.

So you have a circular motion which is translated in space at a constant velocity $\Rightarrow$ helical motion.

The x and y components form a rotating vector.

To solve the differential equation differential differentiate with respect to time the x component equation for $\dfrac{dv_x}{dt}$ again and then substitute for $\dfrac{dv_y}{dt}$ into the y-component equation. You will then get the standard shm differential equation.


A magnetic field can't do any work on a charged particle, so the kinetic energy, and hence its speed, must remain constant.

The Lorentz force acts at right angles to the velocity component perpendicular to the magnetic field. There is no force due to the component of velocity parallel to the magnetic field, since the vector product is zero.

Therefore there is circular motion around the field lines with constant speed and any component of the velocity along the field lines remains constant


I have seen Anedars answer and it looks beautiful. Anyhow I have an objection to your question, which is too long for a comment.

When an electron moves in uniform external magnetic field ... the magnitude of force is ... perpendicular to both components of velocity. But this force, though only due to the perpendicular component, acts on the particle.

If the magnetic field would apply a force to the charged particle, this field has to weaken by time or has to be energized. That this is not the case we can see clearly because the use of a permanent magnet is possible and this magnet does not weaken nor needs to be energized. The Lorentz force powers only from the impuls of the particle (its kinetic energy multiplied with its mass) and its magnetic dipole moment.

The first statement is obvious. It is observed that the particle runs in a spiral path until it stops. Heavier than electrons particles like myons run in a wider spiral path.

The second statement is clear too. Running a positron or a proton, their spiral path is anti-clockwise to the electrons path. This has to do with the intrinsic spin and the related magnetic dipole moment of the particles. If electrons have the magnetic dipole moment parallel to the spin diection of the intrinsic spin, whan the positron has anti-parallel directions for its magnetic moment and the related intrinsic spin. Never something else was observed. And all ouer devices like electric generator, electric engine and electromagnets are working in the same way.

How works this phenomenon in detail:

  • The external magnetic field acts on the magnetic dipole moment and tries to align it.
  • The intrinsic spin of the particle leads to a precession of the particle (gyroscopic effect)
  • Any precession leads to a deflection perpendicular axis of precession and the force, acting on the particle. This is what we observe as the Lorentz force.

That's not all. A one time deflection does not lead to a spiral path. Furthermore any deflection of a particle is an acceleration and due to our knowledge leads to the emission of EM radiation. So the last steps of the phenomenon of Lorentz force are:

  • Once deflected the electron emits a photon and due to the photons impuls the electron's magnetic dipole moment gets misaligned again and the energy, transmitted through the magnetic force of the external magnetic field goes back to this field. The external magnetic field works like a spring. And a small amount of the electrons kinetic energy gets transformed in photons.
  • The process starts again and so on.

So why I added this statement to your question? Because the external magnetic field does not do the work, it is only something like a spring. And second, because the reason for the the spiral path (which in reality has to be a path of tangerine slices) perpendicular to both the movement of the particle and the external field is declared (avowed).


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