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.
Best Answer
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 $\vec{v}(0)$.
Edit:
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. ;-)