PROCEEDINGS OF THE TOMSK POLYTECHNIC INSTITUTE
AWARDED THE ORDER OF THE OCTOBER REVOLUTION
AND THE ORDER OF THE RED BANNER OF LABOR

Volume 206　　　　　　　　　　　　　　　　　　　　　　　　　　　　　　　　1973

FOCUSING OF A CHARGED-PARTICLE BEAM BY A BENDING MAGNET WITH PARALLEL ENTRANCE AND EXIT BOUNDARIES

Yu. P. Kharitonenko, A. I. Komov, N. V. Pirogov

(Presented by the scientific seminar of the “Cyclotron” facility)

Up to the present time, at the cyclotron of the Scientific Research Institute of Nuclear Physics, a sector magnet of the SP-45 type with an aperture angle of \(19^\circ\) has been used to bend the extracted beam. Its principal disadvantages are:

1. Strong horizontal aberration caused by the inequality of the trajectory lengths of the outer ions in the magnetic field.

Figure 1

1. A strong approach of the focus [1].
2. The impossibility of obtaining a point focus when the beam bending angles are not equal to the sector angle of the magnet.

A bending magnet with parallel entrance and exit boundaries is, to a considerable extent, free of these disadvantages. Below we give the calculation of the parameters of a beam that has passed through such a magnet. We shall assume that the magnet does not affect the motion of the ions in the vertical plane, and we shall consider the motion of the ions in the median plane (Fig. 1).

Let a beam of ions with aperture \(2x_{\mathrm{in}}\) and convergence angle \(2\alpha_{\mathrm{in}}\) enter the bending magnet. The beam axis is perpendicular to the entrance boundary of the magnet. In the magnet gap a field of strength

Ions with energy \(E\) move in the field of a magnet along an arc \(\varphi\) with radius \(\rho\), determined [2] from the relation:

\[ H\rho=\frac{1}{0.3z}\sqrt{E^2+2EE_0}, \]

where \(z\) is the ratio of the ion charge to the electron charge;
\(E_0\) is the rest energy of the ion \((M eV)\);
\(H\rho\) is the magnetic “rigidity” of the particle \([kG \cdot cm]\).

To determine the aperture of the beam at the exit of the bending magnet, let us turn to Fig. 2, where \(x_{\text{out}}=cf\) (compare with Fig. 1). Since \(\alpha_{\text{in}}\ll \varphi\), we replace the arcs \(o'o\) and \(gd\) by the corresponding chords, and regard the angles \(ao'o\), \(aoo'\), \(fdg\) as straight. Then from \(\triangle fdg\) we have:

\[ x_{\text{out}}=cf=gf\cos\varphi-cd. \tag{1} \]

The chord \(o'o\simeq \widehat{o'o}=\rho\alpha_{\text{in}}\). From the right triangle \(\triangle obo'\),

\[ o'b=o'o\sin\varphi=\rho\alpha_{\text{in}}\sin\varphi. \]

Assuming further that \(be\simeq oc\), we have \(o'b=cd\). Now, taking (1) into account, we obtain the dependence

\[ x_{\text{out}}=x_{\text{in}}\cos\varphi-\rho\alpha_{\text{in}}\sin\varphi, \tag{2} \]

which relates the beam aperture at the magnet exit to the input beam parameters and the parameters of the magnet.

Fig. 2

To derive the angle \(\alpha_{\text{out}}\) (Fig. 1), let us turn to Fig. 3, where \(\widehat{ab}\) is the trajectory of an ion entering the magnet at a right angle, \(\widehat{ad}\) is the trajectory of an ion having an entrance angle \(\alpha_{\text{in}}\), \(ao=\rho\) is the entrance boundary of the magnet, and \(bd\) is the exit boundary. From Fig. 3 it is seen that the exit angle is the angle between the tangents drawn to the ion trajectories at points \(b\) and \(d\). It is also obvious that the angle between the tangents at points \(b\) and \(d\) is equal to the entrance angle, and consequently the angle between the tangents at points \(c\) and \(d\) is equal to the difference between the exit and entrance angles and is equal to the angle \(co'd\).

Fig. 3

Since \(co'=do'=r\), \(\Delta\alpha\) will be determined after finding the length of the arc \(\widehat{cd}\).

The arc \(\widehat{ab}=\varphi\) (by assumption) and is equal to the arc \(\widehat{ac}\) (by construction). Under the assumption \(\alpha \ll \varphi\), we replace the arc \(\widehat{bc}\) by a chord. And since \(\angle bac=\alpha_{\text{in}}\), while the chords \(ab=ac=2\rho\sin\varphi/2\), then

\[ bc=2\rho\alpha_{\text{in}}\sin\varphi/2 . \tag{3} \]

\(\angle cbd=\angle bae\) (the sides are mutually perpendicular), but \(\angle bae\) (the angle between the tangent and the chord) is equal to half the arc \(\varphi\). Hence \(\angle cbd=\varphi/2\). Taking (3) into account, we have:

\[ cf=2\rho\alpha_{\text{in}}\sin^2\varphi/2, \]

\[ \angle cdb \simeq 90^\circ-\varphi-\alpha_{\text{in}}, \]

therefore

\[ cd=2\rho\alpha_{\text{in}} \frac{\sin^2\varphi/2}{\sin(90^\circ-\varphi-\alpha_{\text{in}})}, \]

and

\[ \Delta x=\frac{cd}{\rho} = 2\alpha_{\text{in}} \frac{\sin^2\varphi/2}{\sin(90^\circ-\varphi-\alpha_{\text{in}})}, \]

and finally,

\[ \alpha_{\text{out}} = \alpha_{\text{in}}+\Delta x = \alpha_{\text{in}} \left[ 1+ 2\frac{\sin^2\varphi/2}{\sin(90^\circ-\varphi-\alpha_{\text{in}})} \right]. \tag{4} \]

Now, from the known aperture of the beam and its convergence angle, it is easy to determine the distance of the focus of the turned beam:

\[ l_n= \frac{x_{\text{out}}}{\alpha_{\text{out}}} = l_{\ell} \frac{ \cos\varphi-\dfrac{\rho}{l_{\ell}}\sin\varphi }{ 1+ 2\dfrac{\sin^2\varphi/2}{\sin(90^\circ-\varphi-\alpha_{\text{in}})} }. \tag{5} \]

In the last equality, the angle \(\alpha_{\text{in}}\) should be taken into account only for \(\varphi\to90^\circ\); in all other cases it may be neglected.

In conclusion we note one remarkable property of the magnet described. If a plane-parallel beam enters it \((\alpha_{\text{in}}=0)\), then, after being turned through any angle \(0<\varphi<90^\circ\), it remains plane-parallel \([\alpha_{\text{out}}=0,\ (4)]\), while its aperture decreases according to (2). From the same formula it is clear that, when a plane-parallel beam is turned through \(\varphi=90^\circ\), the beam aperture in the plane of deflection decreases to zero.

REFERENCES

1. Experimental Nuclear Physics, ed. by E. Segrè, vol. 1, IL, Moscow, 1955.
2. J. Livingood, Principles of Operation of Cyclic Accelerators, IL, 1963.