IPhO 1975 · Theory problem 3

Ions in a magnetic field

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A beam of positive ions (charge \(+e\)) of the same and constant mass \(m\) spread from point Q in different directions in the plane of paper (see figure²). The ions were accelerated by a voltage \(U\). They are deflected in a uniform magnetic field \(B\) that is perpendicular to the plane of paper. The boundaries of the magnetic field are made in a way that the initially diverging ions are focused in point A (\(\overline{QA}=2\cdot a\)). The trajectories of the ions are symmetric to the middle perpendicular on \(\overline{QA}\).

Figure 1

² Remark: This illustrative figure was not part of the original problem formulation.

Among different possible boundaries of magnetic fields a specific type shall be considered in which a contiguous magnetic field acts around the middle perpendicular and in which the points Q and A are in the field free area.

a) Describe the radius curvature \(R\) of the particle path in the magnetic field as a function of the voltage \(U\) and the induction \(B\).

b) Describe the characteristic properties of the particle paths in the setup mentioned above.

c) Obtain the boundaries of the magnetic field boundaries by geometrical constructions for the cases \(R<a\), \(R=a\) and \(R>0\).

d) Describe the general equation for the boundaries of the magnetic field.

Official English text of the problem.

Answers

Take the midpoint of \(QA\) as the origin, with \(Q=(-a,0)\), \(A=(a,0)\), and the illustrated beam above the \(x\)-axis.

  • a)
    \[ \boxed{R=\frac{mv}{eB}=\frac{1}{B}\sqrt{\frac{2mU}{e}}.} \]

  • b) Each trajectory consists of a straight incoming segment, a circular arc of radius \(R\), and a straight outgoing segment. The circle’s centre lies on the perpendicular bisector of \(QA\). If the initial direction makes an angle \(\theta\) with \(QA\), the magnetic deflection is \(2\theta\), and each straight segment has length \[ \boxed{\ell=\frac{a-R\sin\theta}{\cos\theta}.} \]

  • c) Construct the circle centre by intersecting the perpendicular bisector with a line parallel to the incoming ray, displaced by \(R\) towards the magnetic-force side. Its perpendicular foot on the ray is the entry point; reflection gives the exit point.

For the forward-going rays: \[ \begin{array}{c|c} \text{Case}&\text{Allowed angles with }\ell>0\\ \hline R<a&0<\theta<\pi/2\\ R=a&0<\theta<\pi/2\\ R>a&0<\theta<\arcsin(a/R) \end{array} \] For \(R<a\), the complete boundary branches have vertical asymptotes \(x=\pm R\). For \(R\ge a\), they have limiting endpoints at \(Q,A\); a finite angular aperture keeps those points strictly outside the field.

  • d) The portions of the boundary traversed by the beam are \[ \boxed{ x=\pm R\sin\theta,\qquad y=(a-R\sin\theta)\tan\theta } \] or, equivalently, \[ \boxed{ y=\frac{|x|(a-|x|)}{\sqrt{R^2-x^2}}, \qquad 0\le |x|<\min(a,R). } \] Their implicit equation is \[ \boxed{(R^2-x^2)y^2=x^2(a-|x|)^2,} \] with the branch restrictions above.

Solution

a) Radius of curvature

Use the nonrelativistic approximation, neglect the ions’ initial kinetic energy before acceleration, and neglect interactions between ions. Acceleration through the potential difference \(U\) gives

\[ \frac12 mv^2=eU, \qquad v=\sqrt{\frac{2eU}{m}}. \]

The magnetic force is perpendicular to the velocity, so it changes the direction of motion without changing the speed. Inside the uniform field it supplies the centripetal force:

\[ evB=\frac{mv^2}{R}. \]

Consequently,

\[ \boxed{ R=\frac{mv}{eB} =\sqrt{\frac{2mU}{eB^2}}. } \]

Here \(B\) denotes the field magnitude. With SI inputs, \(R\) is in metres.

b) Geometry of the trajectories

Treat the field boundary as sharp, neglecting fringe fields. Let \(O\) be the midpoint of \(QA\), and choose

\[ Q=(-a,0),\qquad A=(a,0). \]

For the outward field direction illustrated, positive ions travelling upwards bend clockwise. Consider an initial direction

\[ \mathbf v=v(\cos\theta,\sin\theta), \qquad 0<\theta<\frac{\pi}{2}. \]

Let \(P\) be the entry point, \(P'\) the exit point, and \(C\) the centre of the circular arc.

Outside the field, the trajectory is straight. At a sharp boundary, the velocity remains continuous, so the straight and circular portions join tangentially. Symmetry about the \(y\)-axis requires \(P'\) to be the reflection of \(P\), and the circle centre must lie on that axis.

Writing \(QP=\ell\),

\[ P=(-a+\ell\cos\theta,\ \ell\sin\theta). \]

The radius from \(P\) towards \(C\) points in the magnetic-force direction:

\[ C-P=R(\sin\theta,-\cos\theta). \]

Since \(C_x=0\),

\[ -a+\ell\cos\theta+R\sin\theta=0, \]

and therefore

\[ \boxed{\ell=\frac{a-R\sin\theta}{\cos\theta}.} \]

It follows that

\[ P=\left(-R\sin\theta,\ (a-R\sin\theta)\tan\theta\right), \]

\[ P'=\left(R\sin\theta,\ (a-R\sin\theta)\tan\theta\right). \]

The centre is

\[ \boxed{ C=\left(0,\frac{a\sin\theta-R}{\cos\theta}\right). } \]

The outgoing velocity has direction \((\cos\theta,-\sin\theta)\). A straight segment of length \(\ell\) then carries the ion from \(P'\) to \(A\), because

\[ P'+\ell(\cos\theta,-\sin\theta)=(a,0). \]

The velocity thus turns through \(2\theta\) inside the field. The circular arc subtends the same angle at \(C\), and its chord has length

\[ PP'=2R\sin\theta. \]

c) Geometrical construction and the three regimes

For a chosen incoming direction, the construction is:

  1. Draw the ray from \(Q\).
  2. Draw a parallel line a perpendicular distance \(R\) away, on the side towards which the magnetic force bends the ion.
  3. Its intersection with the perpendicular bisector of \(QA\) is \(C\).
  4. Drop the perpendicular from \(C\) to the incoming ray. Its foot is \(P\).
  5. Reflect \(P\) in the perpendicular bisector to obtain \(P'\), and draw the clockwise circular arc of radius \(R\) joining them.

Varying the incoming direction traces the entrance boundary. Reflection traces the exit boundary. The perpendicular foot must lie ahead of \(Q\), so

\[ \ell>0 \quad\Longleftrightarrow\quad R\sin\theta<a. \]

For describing the shapes, put \(s=|x|\).

  • \(R<a\).
    Every acute incoming angle is admissible. As \(\theta\) increases from \(0\) towards \(\pi/2\), the boundary runs from \(O\) towards infinite height, approaching \[ \boxed{x=-R\quad\text{and}\quad x=R} \] on the two sides. Since \(a>R\), \(Q\) and \(A\) lie outside this strip.

Near \(O\), \[ y\sim\frac{a}{R}|x|, \] so the complete limiting locus has a pointed bottom.

The branches need not rise monotonically throughout. If \[ \frac{R}{a}>\sqrt{\frac{27}{32}}, \] each branch has a local maximum followed by a local minimum before rising towards its asymptote. Otherwise it rises monotonically. These features follow from the derivative given in part (d).

  • \(R=a\).
    The construction works for every acute angle, and \[ y=s\sqrt{\frac{a-s}{a+s}}, \qquad 0\le s<a. \] Each branch starts at \(O\), rises to a maximum, and returns towards the corresponding source or image point as \(\theta\to\pi/2\).

The maximum occurs at \[ s=\frac{\sqrt5-1}{2}\,a, \] with height \[ y_{\max} =a\left(\frac{\sqrt5-1}{2}\right)^{5/2} \approx0.3003a. \]

The limiting circular trajectory is the upper semicircle of radius \(a\) centred at \(O\), with endpoints \(Q,A\).

  • \(R>a\) — the likely intended third case.
    The construction requires \[ 0<\theta<\alpha,\qquad \boxed{\alpha=\arcsin(a/R)}. \] Geometrically, \(\alpha\) is obtained from a right triangle with hypotenuse \(R\) and opposite side \(a\).

Each boundary branch starts at \(O\), rises to one maximum, and returns towards \(Q\) or \(A\) as \(\theta\to\alpha\). At that limit, \(\ell\to0\), and the circular trajectory has centre \[ C=\left(0,-\sqrt{R^2-a^2}\right). \] Its highest point is \[ y=R-\sqrt{R^2-a^2}. \]

For \(R\ge a\), using the complete limiting locus would put \(Q,A\) on the field boundary. To keep them strictly within field-free areas, choose a beam whose maximum angle is strictly below the limiting value. Continue the unused boundary around its trajectories; for example, close the field above them with a circular arc constructed for a slightly larger admissible angle. The resulting field is connected and remains separated from \(Q,A\).

Only boundary portions actually crossed by the beam are fixed by the focusing condition.

d) General boundary equation

From part (b),

\[ s=R\sin\theta, \qquad y=(a-s)\tan\theta. \]

For the acute-angle branch,

\[ \tan\theta=\frac{s}{\sqrt{R^2-s^2}}. \]

Eliminating \(\theta\) gives

\[ \boxed{ y=f(s)=\frac{s(a-s)}{\sqrt{R^2-s^2}}, \qquad s=|x|,\quad 0\le s<\min(a,R). } \]

Hence

\[ \boxed{(R^2-x^2)y^2=x^2(a-|x|)^2.} \]

The unsquared expression and domain select the physical upper boundary. Squaring alone also admits other branches.

For completeness, differentiating gives

\[ f'(s)= \frac{s^3-2R^2s+aR^2}{(R^2-s^2)^{3/2}}. \]

Thus all stationary heights mentioned in part (c) satisfy

\[ \boxed{s^3-2R^2s+aR^2=0.} \]

For \(R>a\), this equation has exactly one root in \(0<s<a\), giving the maximum of each branch.

We can also verify that the constructed boundary keeps each trajectory in the appropriate region. Write the centre height for entry coordinate \(s\) as

\[ c(s)=f(s)-\sqrt{R^2-s^2} =\frac{as-R^2}{\sqrt{R^2-s^2}}. \]

Then

\[ c'(s)= \frac{R^2(a-s)}{(R^2-s^2)^{3/2}}>0. \]

An ion entering at \(s=s_0\) follows the upper circular arc

\[ y_{\rm arc}(s)=c(s_0)+\sqrt{R^2-s^2}, \qquad 0\le s\le s_0. \]

For \(s<s_0\), monotonicity of \(c(s)\) gives

\[ y_{\rm arc}(s)>c(s)+\sqrt{R^2-s^2}=f(s). \]

Thus the circular portion lies above the entrance and exit boundaries, inside the field. Before entry, the incoming line lies below the boundary because

\[ y_{\rm line}(s)=(a-s)\tan\theta < (a-s)\frac{s}{\sqrt{R^2-s^2}} =f(s) \]

for \(s>s_0\) in the boundary’s domain. Reflection establishes the same result after exit. The boundary therefore produces the required focusing.

The solution was checked against the official solution.