Irodov 3.389: A Velocity Selector Sets the Push – Amit & Deepali Sinha Foundation

Problem Irodov 3.389

A beam of non-relativistic protons travels undeflected through a region of uniform electric and magnetic fields, $E = 120$ kV/m and $B = 50$ mT, that are perpendicular to each other and to the beam. Beyond the fields the beam strikes a grounded target. The beam current is $I = 0.80$ mA. Find the force the beam exerts on the target.

The proton beam crossing the uniform fields E and B, perpendicular to each other and to the beam, without deviation, and striking the target beyond them. — The proton beam crossing the uniform fields $\vect E$ and $\vect B$ , perpendicular to each other and to the beam, without deviation, and striking the target beyond them.

Given

$E = 120$ kV/m
$B = 50$ mT
$I = 0.80$ mA
protons undeviated

Find

the force

F

of the beam on the target

Solution

Step 1The speed the crossed fields select

A proton passes undeviated only if the electric and magnetic forces on it cancel; for $\vect v$ perpendicular to $\vect B$ their magnitudes are $eE$ and $evB$ :

On a proton moving undeviated the electric force q E and the magnetic force q v· B are equal and opposite. — On a proton moving undeviated the electric force $q\vect E$ and the magnetic force $q\vect v\times\vect B$ are equal and opposite.

eE = evB

v = \frac{E}{B}

(1)

With the given fields,

v = \frac{120\times10^{3}\text{ V/m}}{0.050\text{ T}} = 2.4\times10^{6}\text{ m/s},

under $1\%$ of $c$ , so the non-relativistic momentum $mv$ applies.

Step 2The momentum delivered per second

The beam current $I$ carries $I/e$ protons per second, each with momentum $mv$ along the beam. The target, grounded so the charge drains away, stops them and receives the momentum flux. Using $v$ from (1):

F = \frac{I}{e}\,m v = \frac{I}{e}\,m\,\frac{E}{B}

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

F = \frac{(0.80\times10^{-3})(1.672\times10^{-27})(120\times10^{3})}{(1.60\times10^{-19})(0.050)} \text{ N}

F = \frac{mEI}{eB} \approx 20\ \mu\text{N}

(2)

directed along the beam.

Check

Units: $mEI/(eB)$ is $\text{kg}\cdot(\text{V/m})\cdot\text{A}/(\text{C}\cdot\text{T})$ ; with $E/B$ a speed (m/s) and $I/e$ a rate (1/s) this is $\text{kg}\cdot\text{m/s}^2 = \text{N}$ . By energy instead of momentum: a proton at $2.4\times10^{6}$ m/s carries $mv^2/2 = 30$ keV, so the beam delivers the power $P = (30\text{ kV})(0.80\text{ mA}) = 24$ W; for a beam stopped in the target the force is $2P/v = 48/(2.4\times10^{6})$ N $= 20\ \mu$ N, the same value.

Remark

The force depends on the beam only through $I$ and the selected speed $E/B$ , not on how the protons were accelerated: the crossed fields fix $v$ . Had the protons bounced back elastically instead of stopping, the momentum change, and the force, would double. Since $F = 2P/v$ , a slow massive beam pushes hardest for a given beam power.