Irodov 3.379: Same Orbit, Proton and Electron – Amit & Deepali Sinha Foundation

Problem Irodov 3.379

A charged particle circles with radius $r = 100$ mm in a uniform magnetic field of induction $B = 10.0$ mT. Find its speed and its period of revolution when the particle is

a non-relativistic proton;
a relativistic electron.

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

Given

$r = 100$ mm
$B = 10.0$ mT
(a) slow proton
(b) fast electron

Find

the speed

v

and the period

T

in each case

Solution

Step 1The orbit fixes the momentum

The magnetic force $e\vect v\times\vect B$ is perpendicular to the velocity, so the speed is constant and the force only turns the momentum $\vect p$ , at the rate $v/r$ . Equating that rate of change of momentum to the force $evB$ :

\frac{pv}{r} = evB

p = eBr

(1)

Step 2(a) The proton

For a non-relativistic proton of mass $m_p$ the momentum is $m_pv$ . Putting this into (1), and dividing the circumference by the speed for the period:

\begin{aligned} m_pv &= eBr, \\ T &= \frac{2\pi r}{v} = 2\pi r\cdot\frac{m_p}{eBr} \end{aligned}

Numerically, with $e = 1.60\times10^{-19}$ C and $m_p = 1.672\times10^{-27}$ kg,

\begin{aligned} v &= \frac{(1.60\times10^{-19})(1.00\times10^{-2})(0.100)}{1.672\times10^{-27}} \text{ m/s}, \\ T &= \frac{2\pi(1.672\times10^{-27})}{(1.60\times10^{-19})(1.00\times10^{-2})} \text{ s} \end{aligned}

\begin{gathered} v = \frac{eBr}{m_p} \approx 96 \text{ km/s} \\ T = \frac{2\pi m_p}{eB} \approx 6.6\ \mu\text{s} \end{gathered}

(2)

The speed is $3\times10^{-4}$ of $c$ , so the non-relativistic treatment is justified.

Step 3(b) The electron’s speed

For a relativistic electron of rest mass $m_e$ the momentum is $m_ev/\sqrt{1 - v^2/c^2}$ . Setting it equal to (1) and solving for $v$ :

\begin{aligned} \frac{m_ev}{\sqrt{1 - v^2/c^2}} &= eBr, \\ m_e^2v^2 &= (eBr)^2\left(1 - \frac{v^2}{c^2}\right), \\ v^2\left[m_e^2 + \left(\frac{eBr}{c}\right)^2\right] &= (eBr)^2, \\ v^2 &= \frac{c^2}{1 + (m_ec/eBr)^2} \end{aligned}

Numerically, with $m_e = 0.911\times10^{-30}$ kg and $eBr = 1.60\times10^{-22}$ kg m/s:

\begin{aligned} \frac{m_ec}{eBr} &= \frac{(0.911\times10^{-30})(3.00\times10^{8})}{1.60\times10^{-22}} = 1.71, \\ v &= \frac{c}{\sqrt{1 + 1.71^2}} = \frac{c}{1.98} \end{aligned}

v = \frac{c}{\sqrt{1 + (m_ec/eBr)^2}} \approx 0.51\,c

(3)

Step 4(b) The electron’s period

The period is still the circumference over the speed; written through $\gamma = (1 - v^2/c^2)^{-1/2}$ it is the Newtonian period with $m_e$ replaced by $\gamma m_e$ :

T = \frac{2\pi r}{v} = \frac{2\pi r\,\gamma m_e}{\gamma m_e v} = \frac{2\pi r\,\gamma m_e}{eBr}

Numerically, with $v = 0.505\,c$ ,

T = \frac{2\pi(0.100)}{0.505\,(3.00\times10^{8})} \text{ s}

T = \frac{2\pi m_e}{eB\sqrt{1 - v^2/c^2}} \approx 4.1 \text{ ns}

(4)

Check

Units: $eBr/m$ is $\text{C}\cdot\text{T}\cdot\text{m}/\text{kg}$ , and with $\text{T} = \text{kg}/(\text{C}\cdot\text{s})$ that is m/s. The electron’s answers satisfy (1): $\gamma = 1/\sqrt{1 - 0.505^2} = 1.16$ , and $\gamma m_ev = 1.16\,(0.911\times10^{-30})(1.52\times10^{8}) = 1.60\times10^{-22}$ kg m/s $= eBr$ . Limit: when $m_ec \gg eBr$ the speed formula reduces to $v = eBr/m_e$ , the form used for the proton.

Remark

Both particles carry the same momentum, $eBr = 1.6\times10^{-22}$ kg m/s, because the orbit and the field are the same; what differs is how much speed that momentum buys. For the heavy proton it is a leisurely 96 km/s. For the electron the Newtonian formula would claim $eBr/m_e = 0.59\,c$ , too fast by 16 per cent; and its period grows with its energy through $\gamma$ , which is why an electron falls out of step with a fixed-frequency cyclotron after very little acceleration.