Irodov 3.380: Momentum Read from the Orbit – Amit & Deepali Sinha Foundation

Problem Irodov 3.380

A relativistic particle of charge $q$ and rest mass $m_0$ moves on a circle of radius $r$ in a uniform magnetic field of induction $B$ . Find:

the magnitude of its momentum;
its kinetic energy;
its acceleration.

The charge q on its circle of radius r in the uniform field B, normal to the plane of the circle (drawn out of the page, for q > 0). — The charge $q$ on its circle of radius $r$ in the uniform field $\vect B$ , normal to the plane of the circle (drawn out of the page, for $q > 0$ ).

Given

$q$
$m_0$
$r$
$B$

Find

(a) the momentum

p

· (b) the kinetic energy

T

· (c) the acceleration

w

Solution

Step 1(a) The momentum from the radius

The force $q\vect v\times\vect B$ is perpendicular to the velocity and does no work, so the speed and the magnitude of the momentum are constant. The force only turns $\vect p$ , at the rate $v/r$ , so the rate of change of the momentum has magnitude $pv/r$ . Equating it to the force $qvB$ :

\frac{pv}{r} = qvB

p = qBr

(1)

Step 2(b) The kinetic energy

The kinetic energy is the total energy less the rest energy, and the total energy follows from the momentum through $W^2 = (pc)^2 + (m_0c^2)^2$ . With (1):

T = \sqrt{(qBrc)^2 + (m_0c^2)^2} - m_0c^2

T = m_0c^2\left[\sqrt{1 + \left(\frac{qBr}{m_0c}\right)^2} - 1\right]

(2)

Step 3The speed

The velocity is $\vect v = \vect pc^2/W$ . Using (1) for $p$ and the same $W$ , then dividing through by $qBr$ :

v = \frac{qBrc^2}{\sqrt{(qBrc)^2 + (m_0c^2)^2}} = \frac{c}{\sqrt{1 + (m_0c/qBr)^2}} \tag{3}

Step 4(c) The acceleration

The speed is constant, so the acceleration is purely centripetal, $v^2/r$ . Squaring (3) and dividing by $r$ :

w = \frac{v^2}{r} = \frac{1}{r}\cdot\frac{c^2}{1 + (m_0c/qBr)^2}

w = \frac{c^2}{r\left[1 + (m_0c/qBr)^2\right]}

(4)

It points to the centre of the circle.

Check

Newtonian limit, $qBr \ll m_0c$ : expanding the root in (b) to first order in $(qBr/m_0c)^2$ , and dropping that term in the bracket of (c),

\begin{gathered} T \to \frac{(qBr)^2}{2m_0} = \frac{p^2}{2m_0}, \\ w \to \frac{(qBr)^2}{m_0^2\,r} \end{gathered}

the kinetic energy of a slow particle, and $v^2/r$ for the speed $qBr/m_0$ of the familiar non-relativistic orbit. Ultra-relativistic limit, $qBr \gg m_0c$ : $T$ tends to $pc$ and $w$ to $c^2/r$ , a particle moving at essentially $c$ . Units: $qBr$ is $\text{C}\cdot\text{T}\cdot\text{m}$ , that is $\text{kg}\cdot\text{m/s}$ , a momentum.

Remark

Result (a) is exact at every energy, which makes a magnetic field the standard momentum meter of particle physics: a curved track of radius $r$ in a field $B$ gives $p = qBr$ directly, and for a singly charged particle this reads $p\,[\text{GeV}/c] \approx 0.3\,B\,[\text{T}]\;r\,[\text{m}]$ . The energy and the speed need the rest mass as well, which is how the same track, with an independent speed measurement, identifies the particle.