Fyneman exercise II ch1-6

Q.

A particle with a mass $m$ and a positive charge $q$ is at a point $x=z=0$, $y=a$, and is moving with a low velocity

$\vec{v}=v_0\hat{e}_x$
The charge is influenced by a negative charge $-Q$ fixed at the origin and by a uniform magnetic field $B_0$ in the $+z$ direction. How large must $B_0$ be such that the moving particle describes a circle of radius $a$ about the stationary one? If the magnitude of the magnetic field strength were different than this, explain why the speed of the particle is a function of the radial distance only.
Sketch qualitatively several cycles of the trajectory followed by the particle if it were released from the point $x=z=0$, $y=a$ with zero velocity.

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Trial.

For circular motion,
$acceleration\ \vec{a}=-{v^2 \over r} \hat{e}_r\tag{1}$
$v=const.\tag{2}$
$a=const\tag{3}$
For our system, if eq(1) is satisfied, it starts circular motion, then at start time radius $r=a$, $v=v_0$and
$m\vec{a}=q(-{Q \over 4\pi\epsilon_0 a^2}\hat{e}_y+v_0\hat{e}_x\times B_0\hat{e}_z)\tag{4}$
From eq(1),
$\vec{a}=-{v_0^2 \over a}\hat{e}_y\tag{5}$
From eq(4),
$\vec{a}=-{q \over m}({Q \over 4\pi\epsilon_0 a^2}+v_0B_0)\hat{e}_y\tag{6}$
From eq(5) & eq(6), condition of $B_0$ for circular motion is
$B_0={mv_0 \over qa}-{Q \over 4\pi\epsilon_0 a^2v_0}\tag{7}$
Motion is always in x-y plane, because force is always toward origin.
That the force is always toward origin means $\Delta \vec{v}$ is influenced only by radial distance and direction is always toward orgin. So, speed is a function of the radial distance only.

sketch