IPhO 1977 · Theory problem 3

Electron beam aimed at a target by a magnetic field

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An electron gun \(T\) emits electrons accelerated by a potential difference \(U\) in a vacuum in the direction of the line \(a\) as shown in Fig. 2. The target \(M\) is placed at a distance \(d\) from the electron gun in such a way that the line segment connecting the points \(T\) and \(M\) and the line \(a\) subtend the angle \(\alpha\) as shown in Fig. 2. Find the magnetic induction \(B\) of the uniform magnetic field

Figure 1

Figure 2:

a) perpendicular to the plane determined by the line \(a\) and the point \(M\)

b) parallel to the segment \(TM\)

in order that the electrons hit the target \(M\). Find first the general solution and then substitute the following values: \(U = 1000\ \mathrm{V}\), \(e = 1.60 \cdot 10^{-19}\ \mathrm{C}\), \(m_e = 9.11 \cdot 10^{-31}\ \mathrm{kg}\), \(\alpha = 60^\circ\), \(d = 5.0\ \mathrm{cm}\), \(B < 0.030\ \mathrm{T}\).

Official English text of the problem.

Answers

  • a) The field must point into the page in Fig. 2, with magnitude \[ \boxed{B=\frac{2\sin\alpha}{d}\sqrt{\frac{2m_eU}{e}} \simeq 3.7\times10^{-3}\ \mathrm{T}.} \]

  • b) Either direction parallel to \(TM\) is possible. The allowed magnitudes are \[ \boxed{B_n=\frac{2\pi n\cos\alpha}{d}\sqrt{\frac{2m_eU}{e}}, \qquad n=1,2,3,\ldots.} \] With the supplied values and \(B<0.030\ \mathrm{T}\), only \(n=1,2,3,4\) are allowed: \[ \boxed{B\simeq 6.7\times10^{-3},\quad 1.3\times10^{-2},\quad 2.0\times10^{-2},\quad 2.7\times10^{-2}\ \mathrm{T}.} \]

Solution

Let \(e>0\) denote the magnitude of the electron charge, so the charge is \(-e\). We use nonrelativistic mechanics, neglect the initial kinetic energy before acceleration, and consider motion in the uniform magnetic field after the electrons leave the gun.

The accelerating potential supplies kinetic energy: \[ \frac12m_ev^2=eU, \qquad v=\sqrt{\frac{2eU}{m_e}}. \] The magnetic force \[ \mathbf F=-e\,\mathbf v\times\mathbf B \] is perpendicular to the velocity and therefore does no work. Consequently, the speed remains constant throughout the trajectory.

a) Field perpendicular to the plane of the figure

The velocity remains perpendicular to the magnetic field. The magnetic force provides the centripetal force for circular motion: \[ evB=\frac{m_ev^2}{R}, \qquad R=\frac{m_ev}{eB}. \]

Take \(T=(0,0)\), with the initial velocity along the positive \(x\)-axis and the positive \(y\)-axis upward in the figure. Then \[ M=(d\cos\alpha,-d\sin\alpha). \] The orbit must bend downward, so its centre is \(C=(0,-R)\). Requiring \(M\) to lie on this circle gives \[ (d\cos\alpha)^2+(R-d\sin\alpha)^2=R^2. \] Thus \[ d^2-2Rd\sin\alpha=0, \qquad R=\frac{d}{2\sin\alpha}. \] Combining the geometrical and dynamical expressions for \(R\), \[ \boxed{B=\frac{2m_ev\sin\alpha}{ed} =\frac{2\sin\alpha}{d}\sqrt{\frac{2m_eU}{e}}.} \]

To determine the direction, note that an electron initially moving to the right must experience a downward force. Because its charge is negative, the field must point into the page.

For the supplied data, \[ v=\sqrt{\frac{2(1.60\times10^{-19})(1000)} {9.11\times10^{-31}}} \simeq1.87\times10^7\ \mathrm{m\,s^{-1}}, \] consistent with the nonrelativistic approximation. The required radius is \[ R=\frac{0.050}{2\sin60^\circ} \simeq2.9\times10^{-2}\ \mathrm{m}, \] and the field is \[ B=3.6966\ldots\times10^{-3}\ \mathrm{T} \simeq\boxed{3.7\times10^{-3}\ \mathrm{T}}. \] This satisfies \(B<0.030\ \mathrm{T}\).

b) Field parallel to \(TM\)

Resolve the initial velocity into components along \(TM\) and perpendicular to it: \[ v_\parallel=v\cos\alpha, \qquad v_\perp=v\sin\alpha. \] The parallel component remains constant because the magnetic force has no component along the field. The perpendicular component produces uniform circular motion, so the complete trajectory is a helix whose axis is parallel to \(TM\).

For the circular motion, \[ ev_\perp B=\frac{m_ev_\perp^2}{r}, \qquad \omega=\frac{v_\perp}{r}=\frac{eB}{m_e}. \] Hence its period is \[ P=\frac{2\pi}{\omega}=\frac{2\pi m_e}{eB}. \]

Since \(T\) and \(M\) differ only in position along the field direction, the electron must return to its initial transverse position when it reaches \(M\). For \(v_\perp\ne0\), this happens precisely after an integer number \(n\) of complete revolutions: \[ t=nP,\qquad n=1,2,3,\ldots. \] During this time, the displacement along \(TM\) must equal \(d\): \[ d=v\cos\alpha\,nP =v\cos\alpha\,\frac{2\pi n m_e}{eB}. \] Therefore, \[ \boxed{B_n=\frac{2\pi n m_ev\cos\alpha}{ed} =\frac{2\pi n\cos\alpha}{d} \sqrt{\frac{2m_eU}{e}}, \qquad n=1,2,3,\ldots.} \]

Reversing the magnetic field reverses the sense of rotation but leaves both the period and the motion along \(TM\) unchanged. Thus both field directions parallel to \(TM\) work.

For the numerical data, the fundamental field magnitude is \[ B_1=\frac{2\pi\cos60^\circ}{0.050} \sqrt{\frac{2(9.11\times10^{-31})(1000)} {1.60\times10^{-19}}} =0.0067049\ldots\ \mathrm{T}. \] The upper bound requires \[ nB_1<0.030\ \mathrm{T}, \qquad n<4.4743\ldots. \] Thus the allowed integers and corresponding fields, rounded to two significant figures, are

Number of complete revolutions \(n\) Magnetic induction \(B_n\)
1 \(6.7\times10^{-3}\ \mathrm{T}\)
2 \(1.3\times10^{-2}\ \mathrm{T}\)
3 \(2.0\times10^{-2}\ \mathrm{T}\)
4 \(2.7\times10^{-2}\ \mathrm{T}\)

The next value, \(B_5\simeq0.0335\ \mathrm{T}\), exceeds the limit.

These discrete solutions apply to the acute, nonzero angle shown. If \(\alpha=0\), the electrons already travel along \(TM\), and any parallel magnetic field leaves that straight motion unchanged. If \(\cos\alpha\le0\), the electrons cannot reach a target a positive distance along \(TM\), because their velocity component in that direction cannot increase.

The solution was checked against the official solution.