Irodov 3.378: Bending Through a Magnetic Slab – Amit & Deepali Sinha Foundation

Problem Irodov 3.378

A proton, accelerated from rest through a potential difference $V = 500$ kV, crosses a layer of uniform magnetic field of induction $B = 0.51$ T and thickness $d = 10$ cm. The proton enters perpendicular to the faces of the layer, and the field is parallel to the faces and perpendicular to the proton’s velocity, pointing into the page of the figure. Find the angle $\alpha$ by which the proton’s direction of motion is turned from its initial direction.

The layer of uniform field B (into the page), thickness d. The proton enters normal to the layer, is turned inside it, and leaves at the angle α to its initial direction. — The layer of uniform field $\vect B$ (into the page), thickness $d$ . The proton enters normal to the layer, is turned inside it, and leaves at the angle $\alpha$ to its initial direction.

Given

$V = 500$ kV
$B = 0.51$ T
$d = 10$ cm
entry normal to the layer

Find

the deflection angle

\alpha

Solution

Step 1The momentum the voltage gives

The proton, of charge $e$ and rest mass $m$ , gains the kinetic energy $eV$ . The exact relation between momentum and kinetic energy gives

\begin{align*} (pc)^2 &= eV\left(eV + 2mc^2\right), \\ p &= \sqrt{2meV\left(1 + \frac{eV}{2mc^2}\right)} \tag{1}\end{align*}

Here $eV = 0.5$ MeV against $mc^2 = 938$ MeV, so the bracket is $1.0003$ .

Step 2A circular arc inside the layer

The magnetic force $e\vect v\times\vect B$ is perpendicular to the velocity, so it does no work: the speed, and with it $p$ , stays constant, and the force only turns the momentum. Turning a momentum $p$ at the rate $v/R$ takes the force $evB$ :

\frac{pv}{R} = evB

R = \frac{p}{eB}

(2)

Step 3The geometry of the crossing

The proton enters along the normal to the layer, so the centre of its arc lies on the entry face, as drawn below.

The crossing with v and the magnetic force F at a point of the arc, and the radius R from the centre, on the entry face, to the exit point. The exit radius is at α to the face, so Rα = d. — The crossing with $\vect v$ and the magnetic force $\vect F$ at a point of the arc, and the radius $R$ from the centre, on the entry face, to the exit point. The exit radius is at $\alpha$ to the face, so $R\sin\alpha = d$ .

On leaving, the radius is perpendicular to the exit velocity, so it makes the angle $\alpha$ with the entry face, and its reach across the layer is the thickness $d$ :

R\sin\alpha = d \tag{3}

This needs $d < R$ ; otherwise the proton turns back inside the layer.

Step 4The deflection angle

Combining (3) with (2) and then (1):

\sin\alpha = \frac{d}{R} = \frac{eBd}{p} = \frac{dB\sqrt{e/2mV}}{\sqrt{1 + eV/2mc^2}}

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

\sin\alpha = \frac{(0.10)(0.51)\sqrt{\dfrac{1.60\times10^{-19}}{2(1.672\times10^{-27})(5.0\times10^{5})}}}{\sqrt{1.0003}} = 0.499

\alpha = \arcsin\frac{dB\sqrt{e/2mV}}{\sqrt{1 + eV/2mc^2}} \approx 30^\circ

(4)

At this energy $eV/2mc^2 = 2.7\times10^{-4}$ , so the root in the denominator is 1 to four figures and $\alpha = \arcsin\bigl(dB\sqrt{e/2mV}\bigr)$ , the non-relativistic form, gives the same $30^\circ$ .

Check

Units: $e/(mV)$ is $\text{C}/(\text{kg}\cdot\text{V})$ ; with $\text{V} = \text{J/C}$ and $\text{T} = \text{kg}/(\text{C}\cdot\text{s})$ , $B^2 e/(mV)$ is $1/\text{m}^2$ , so $dB\sqrt{e/mV}$ is a pure number, as a sine must be. Directly: $p = 1.64\times10^{-20}$ kg m/s, so $R = p/(eB) = 0.20$ m $= 2d$ , and $\sin\alpha = 1/2$ .

Remark

The layer is a momentum filter. A proton with $p < eBd$ has $R < d$ , never reaches the far face, and comes back out of the entry face after half a turn; only momenta above $eBd$ get through, each deflected by its own angle $\arcsin(eBd/p)$ . Spread over a detector, the exit angles sort a beam by momentum, the principle of a magnetic spectrometer.