In Cartesian coordinates the Lagrangian is
$$
\mathcal L=\frac12(\dot x^2+\dot y^2+\dot z^2)+y\dot x.
$$
Introduce a <Lagrange multiplier> $\lambda$ for $x^2+y^2+z^2=c^{-2}$. The <Euler-Lagrange equations> become
$$
\ddot x+\dot y=\lambda x,
\qquad
\ddot y-\dot x=\lambda y,
\qquad
\ddot z=\lambda z,
$$
or
$$
\ddot{\mathbf x}+\dot y\,\mathbf i-\dot x\,\mathbf j
=\lambda\mathbf x.
$$
Differentiating the constraint twice gives
$$
\mathbf x\mathbin\cdot\dot{\mathbf x}=0,
\qquad
\mathbf x\mathbin\cdot\ddot{\mathbf x}=-|\dot{\mathbf x}|^2.
$$
Taking the scalar product of the vector equation with $\mathbf x$ therefore yields
$$
\lambda=c^2\left(-|\dot{\mathbf x}|^2+x\dot y-y\dot x\right).
$$
Substitution gives
$$
\boxed{\ddot{\mathbf x}+\dot y\,\mathbf i-\dot x\,\mathbf j
+c^2\left(|\dot{\mathbf x}|^2+y\dot x-x\dot y\right)\mathbf x=0.}
$$

For $c=0$, the equations have the first integrals
$$
\dot x+y=A,
\qquad
\dot y-x=B,
\qquad
\dot z=C.
$$
Writing $X=x+B$ and $Y=y-A$ gives $\dot X=-Y$ and $\dot Y=X$. Hence every trajectory is
$$
\boxed{x(t)=R\cos(t-t_0)-B,
\qquad
y(t)=R\sin(t-t_0)+A,
\qquad
z(t)=Ct+D.}
$$
These are <helical motion in a uniform magnetic field>[helices] of unit angular frequency about a line parallel to the $z$ axis, including circles when $C=0$ and a straight line when $R=0$.

Solution

Past exam of the mathematics course of the University of Cambridge / 2018 / ii / Paper 2 / 8B / Solution

The <Lorentz force> and <Newton's second law> give
$$
m\ddot{\mathbf x}=q\dot{\mathbf x}\times\mathbf B.
$$
Writing $\Omega=qB/m$, the component equations are
$$
\ddot x=\Omega\dot y,\qquad \ddot y=-\Omega\dot x,\qquad \ddot z=0.
$$
The initial <velocity> therefore gives
$$
\dot x=v\sin(\Omega t),\qquad \dot y=v\cos(\Omega t),\qquad \dot z=v,
$$
and integration with the initial <position> gives
$$
\boxed{\mathbf x(t)=\left(1+\frac v\Omega[1-\cos(\Omega t)],\ \frac v\Omega\sin(\Omega t),\ vt\right)}.
$$
This is <helical motion in a uniform magnetic field>: a <helix> of radius $v/\Omega$ around the line $x=1+v/\Omega$, $y=0$, with its axis parallel to the <magnetic field>.

\Image[/media/past-exam-of-the-mathematics-course-of-the-university-of-cambridge/2019/ia/paper-4-magnetic-helix.png]
{height=520}

Past exam of the mathematics course of the University of Cambridge / 2019 / ia / Paper 4 / 10A / a / Solution

In a uniform <magnetic field>, the <Lorentz force> leaves the component of <velocity> parallel to the field constant and rotates the perpendicular component at the nonrelativistic <cyclotron frequency> $|q|B/m$. The resulting path is a <helix> around a field line.

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