Past exam of the mathematics course of the University of Cambridge / 2010 / ia / Paper 4 / 3B / Solution

With no electric field, the Lorentz force is $F=q\dot{r}\times B\hat{z}$. The equation from Newton's second law and its first integral areThe cyclotron frequency is $|\omega |$; the sign of $\omega$ specifies the sense of rotation. Since $\hat{x}\times \hat{z}=-\hat{y}$, the initial data giveThe component equations are $\dot{x}=\omega y$, $\dot{y}+\omega x=u+a\omega$ and $\dot{z}=v$. For $\omega \ne 0$, differentiating the first givesSolving this linear differential equation and then using $y=\dot{x}/\omega$ givesThe perpendicular projection is a circle of radius $|u/\omega |$ centred at $(a+u/\omega ,0)$, the guiding centre. The velocity parallel to the magnetic field stays constant, producing helical motion in a uniform magnetic field.

When $a\omega +u=0$, this simplifies toThus the trajectory is a helix about the $z$ axis, with radius $|a|$ and axial advance $2\pi v/|\omega |$ per complete revolution in increasing time. Its projection moves clockwise when viewed from positive $z$ for $\omega >0$. If $v=0$ it is a circle; if $a=0$ it is a straight line along the axis.

For $\omega =0$, the Lorentz force vanishes and the solution is instead $r(t)=a\hat{x}+ut\hat{y}+vt\hat{z}$, also the limit of the general formula. The special condition then forces $u=0$.