Irodov 3.407: Path Length Is Energy over Force – Amit & Deepali Sinha Foundation

Problem Irodov 3.407

In a betatron the electrons circulate on an equilibrium orbit of radius $r$ . During the acceleration the magnetic induction at the orbit rises at an essentially constant rate from zero to $B$ . An electron starts from rest. Find:

the energy the electron acquires during the acceleration;
the distance it travels, if the acceleration lasts a time $\Delta t$ .

A betatron seen along its magnetic field B (drawn out of the page; schematically, its glyphs are denser inside the orbit, where the field is stronger): the electron on the equilibrium orbit of radius r. — A betatron seen along its magnetic field $\vect B$ (drawn out of the page; schematically, its glyphs are denser inside the orbit, where the field is stronger): the electron on the equilibrium orbit of radius $r$ .

Given

$r$
$B$ on the orbit: $0 \to B$ at a constant rate
start from rest
acceleration time $\Delta t$

Find

(a) the energy

W

acquired · (b) the distance

s

covered

Solution

Step 1The momentum at every moment

On the orbit the magnetic force $e v B_0$ , with $B_0$ the induction there, provides the rate $p v/r$ at which the momentum turns. Equating the two and cancelling $v$ , at any speed:

\frac{p v}{r} = e v B_0 \quad\Longrightarrow\quad p = e r B_0 \tag{1}

At the end of the acceleration $B_0 = B$ , so the final momentum is $erB$ .

Step 2(a) The energy acquired

The total energy $m_0c^2 + W$ , rest energy plus kinetic, follows from the momentum by $(m_0c^2 + W)^2 = p^2c^2 + m_0^2c^4$ . Solving for $W$ with the final momentum from (1):

W = \sqrt{(erB)^2c^2 + m_0^2c^4} - m_0c^2

Taking $m_0c^2$ out of the root:

W = m_0c^2\left(\sqrt{1 + \left(\frac{erB}{m_0c}\right)^2} - 1\right)

(2)

Step 3The momentum grows linearly in time

The induction on the orbit rises at a constant rate, $B_0 = B\,t/\Delta t$ , so by (1) the momentum grows uniformly from zero:

p(t) = k\,t, \qquad k = \frac{erB}{\Delta t} \tag{3}

The constant $k = dp/dt$ is the tangential force $eE$ of the eddy field, the same throughout the acceleration.

Step 4(b) The distance covered

The speed follows from the momentum and the total energy:

v = \frac{pc^2}{\sqrt{p^2c^2 + m_0^2c^4}} = \frac{pc}{\sqrt{p^2 + m_0^2c^2}}

Integrating it over the acceleration time and changing the variable to $p$ with (3), $dt = dp/k$ :

\begin{aligned} s &= \int_0^{\Delta t} v\,dt = \frac{c}{k}\int_0^{erB}\frac{p\,dp}{\sqrt{p^2 + m_0^2c^2}} \\ &= \frac{c}{k}\left(\sqrt{(erB)^2 + m_0^2c^2} - m_0c\right) \end{aligned}

The bracket times $c$ is $\sqrt{(erB)^2c^2 + m_0^2c^4} - m_0c^2$ , which is the energy $W$ of (2), so the path is that energy over the constant force:

s = \frac{W}{k}

With $k$ from (3):

s = \frac{W\Delta t}{erB}

(4)

Check

The result $s = W/k$ is the work-energy theorem: the eddy field’s force $k$ is constant and always along the velocity, so its work over the path is $ks$ , which must equal $W$ . Limits: for a slow electron $W = (erB)^2/2m_0$ , and $s = (erB/m_0)\,\Delta t/2$ , the final speed times $\Delta t/2$ , as for uniform acceleration from rest; for an ultrarelativistic one $W \approx erBc$ and $s \approx c\Delta t$ , light speed for the whole time. Units: $erB$ is a momentum, and an energy over a momentum is a speed, times $\Delta t$ a length.

Remark

The integral was not needed: because the driving force is constant and tangential, the distance is simply the energy gained divided by that force, at any speed, relativistic or not, and whatever the speed does along the way. The same argument gives the number of turns, $s/2\pi r = W\Delta t/(2\pi er^2B)$ , which is $W$ divided by the energy per turn $2\pi er^2B/\Delta t$ .