IPhO 2000 · Theory problem 2

Charge-to-mass ratio of the electron (magnetic focusing in a CRT; beta particles between charged plates)

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(a) A cathode ray tube (CRT), consisting of an electron gun and a screen, is placed within a uniform constant magnetic field of magnitude \(\mathbf B\) such that the magnetic field is parallel to the beam axis of the gun, as shown in figure 2.3.

Figure 2.3

Figure 2.3

The electron beam emerges from the anode of the electron gun on the axis, but with a divergence of up to \(5^\circ\) from the axis, as illustrated in figure 2.4. In general a diffuse spot is produced on the screen, but for certain values of the magnetic field a sharply focused spot is obtained.

Figure 2.4

Figure 2.4

By considering the motion of an electron initially moving at an angle \(\beta\) (where \(0\leq\beta\leq5^\circ\)) to the axis as it leaves the electron gun, and considering the components of its motion parallel and perpendicular to the axis, derive an expression for the charge to mass ratio \(e/m\) for the electron in terms of the following quantities:

the smallest magnetic field for which a focused spot is obtained,
the accelerating potential difference across the electron gun \(V\)
(note that \(V<2\ \mathrm{kV}\)),
and \(D\), the distance between the anode and the screen.

Write your expression in the box provided in section 2a of the answer sheet.

(b) Consider another method of evaluating the charge to mass ratio of the electron. The arrangement is shown from a side view and in plan view (from above) in figure 2.5, with the direction of the magnetic field marked \(\mathbf B\). Within this uniform magnetic field \(\mathbf B\) are placed two brass circular plates of radius \(\rho\) which are separated by a very small distance \(t\). A potential difference \(V\) is maintained between them. The plates are mutually parallel and co-axial, however their axis is perpendicular to the magnetic field. A photographic film, covers the inside of the curved surface of a cylinder of radius \(\rho+s\), which is held co-axial with the plates. In other words, the film is at a radial distance \(s\) from the edges of the plates. The entire arrangement is placed in vacuo. Note that \(t\) is very much smaller than both \(s\) and \(\rho\).

Figure 2.5

Figure 2.5

A point source of \(\beta\) particles, which emits the \(\beta\) particles uniformly in all directions with a range of velocities, is placed between the centres of the plates, and the same piece of film is exposed under three different conditions:

firstly with \(B=0\), and \(V=0\),
secondly with \(B=B_0\), and \(V=V_0\), and
thirdly with \(B=-B_0\), and \(V=-V_0\);

where \(V_0\) and \(B_0\) are positive constants. Please note that the upper plate is positively charged when \(V>0\) (negative when \(V<0\)), and that the magnetic field is in the direction defined by figure 2.5 when \(B>0\) (in the opposite direction when \(B<0\)). For this part you may assume the gap is negligibly small.

Two regions of the film are labelled A and B on figure 2.5. After exposure and development, a sketch of one of these regions is given in figure 2.6. From which region was this piece taken (on your answer sheet write A or B)? Justify your answer by showing the directions of the forces acting on the electron.

Figure 2.6

Figure 2.6

(c) After exposure and development, a sketch of the film is given in figure 2.6. Measurements are made of the separation of the two outermost traces with a microscope, and this distance (\(y\)) is also indicated for one particular angle on figure 2.6. The results are given in the table below, the angle \(\phi\) being defined in figure 2.5 as the angle between the magnetic field and a line joining the centre of the plates to the point on the film.

Angle to field /degrees \(\phi\) 90 60 50 40 30 23
Separation /mm \(y\) 17.4 12.7 9.7 6.4 3.3 End of trace

Numerical values of the system parameters are given below:

\[ B_0=6.91\ \mathrm{mT} \qquad V_0=580\ \mathrm V \qquad t=0.80\ \mathrm{mm} \] \[ s=41.0\ \mathrm{mm} \]

In addition, you may take the speed of light in vacuum to be \(3.00\times10^8\ \mathrm{m\,s^{-1}}\), and the rest mass of the electron to be \(9.11\times10^{-31}\ \mathrm{kg}\).

Determine the maximum \(\beta\) particle kinetic energy observed.

Write the maximum kinetic energy as a numerical result in eV in the box on the answer sheet, section 2f.

(d) Using the information given in part (f) [i.e. part (c) above], obtain a value for the charge to rest mass ratio of the electron. This should be done by plotting an appropriate graph on the paper provided.

Indicate algebraically the quantities being plotted on the horizontal and vertical axes both on the graph itself and on the answer sheet in the boxes provided in section 2g.

Write your value for the charge to mass ratio of the electron in the box provided on the answer sheet, section 2g.

Please note that the answer you obtain may not agree with the accepted value because of a systematic error in the observations.

Official English text of the problem.

Answers

  • (a) For the smallest positive focusing field \(B_{\min}\), \[ \boxed{\frac em\simeq\frac{8\pi^2V}{B_{\min}^{\,2}D^2}}. \]

  • (b) \(\boxed{\text{Region B}}\). For \(B=B_0,\ V=V_0\), the electric force is upward and the magnetic force downward. Reversing both fields reverses both forces.

  • (c) \[ \boxed{K_{\max}\simeq6.4\times10^5\ \mathrm{eV}} \] corresponding to \(v_{\max}\simeq2.69\times10^8\ \mathrm{m\,s^{-1}}\).

  • (d) Plot \(Y=y\) vertically against the dimensionless horizontal quantity \[ X=\sin\phi\sqrt{\sin^2\phi- \left(\frac{V_0}{B_0tc}\right)^2}. \] The straight line has slope approximately \(1.855\times10^{-2}\ \mathrm m\), giving \[ \boxed{\frac{e}{m_0}\simeq1.68\times10^{11}\ \mathrm{C\,kg^{-1}}}. \]

Solution

Throughout, \(e>0\) denotes the magnitude of the electron charge; the electron’s charge is \(-e\). Its rest mass is \(m_0\).

(a)

Since the accelerating voltage is below \(2\ \mathrm{kV}\), the electron’s kinetic energy is much smaller than its rest energy. We use nonrelativistic mechanics, with \(m\simeq m_0\):

\[ eV=\frac12mv^2. \]

The velocity components on leaving the gun are \[ v_\parallel=v\cos\beta,\qquad v_\perp=v\sin\beta. \]

The magnetic field exerts no force parallel to itself. The parallel motion is therefore uniform, while the perpendicular motion is circular, with \[ r=\frac{mv_\perp}{eB}, \qquad \omega=\frac{eB}{m}, \qquad T=\frac{2\pi m}{eB}. \]

The resulting trajectory is a helix. An electron starting on the axis returns to the axis after each complete revolution. The smallest positive focusing field corresponds to one revolution before reaching the screen, so \[ D=v\cos\beta\,T =\frac{2\pi mv\cos\beta}{eB_{\min}}. \]

Squaring and substituting \(v^2=2eV/m\) gives \[ \frac em =\frac{8\pi^2V\cos^2\beta}{B_{\min}^{\,2}D^2}. \]

For the narrow beam, \(\beta\leq5^\circ\), we take \(\cos\beta\simeq1\). Hence \[ \boxed{\frac em\simeq\frac{8\pi^2V}{B_{\min}^{\,2}D^2}}. \]

The focusing is approximate for a beam containing different \(\beta\): their axial velocities differ slightly. At \(5^\circ\), \(1-\cos\beta\) is only about \(0.0038\).

(b)

The Lorentz force is \[ \mathbf F=-e\left(\mathbf E+\mathbf v\times\mathbf B\right). \]

For \(V=V_0>0\), the upper plate is positive, so the electric field points downward and the electric force on an electron points upward.

For \(B=B_0>0\), applying the cross-product rule to the plan view gives:

Electron travelling toward Electric force Magnetic force
Region A Upward Upward
Region B Upward Downward

With a negligibly small gap, electrons can traverse the plates only when their vertical forces balance. This is possible in region B.

Let \(\theta\) be the angle between the electron’s velocity and the magnetic field while it travels between the plates. Neglecting fringing, \[ E_0=\frac{V_0}{t}. \] Force balance requires \[ eE_0=eB_0v\sin\theta, \] or \[ \boxed{v\sin\theta=\frac{V_0}{B_0t}}. \]

Reversing both \(B\) and \(V\) reverses both forces, so they still balance for electrons travelling toward region B.

After an electron leaves the plates, the electric force disappears while the magnetic force remains. Thus:

  • \(B=B_0,\ V=V_0\) produces the lower curved trace.
  • \(B=-B_0,\ V=-V_0\) produces the upper curved trace.
  • \(B=V=0\) produces the central straight trace.

These are the three traces shown in figure 2.6, establishing \[ \boxed{\text{Region B}}. \]

(c)

We use the thin-gap approximation and, to leading order in the subsequent deflection, identify the electron’s departure angle \(\theta\) with the observed angle \(\phi\). Define \[ u=\frac{V_0}{B_0t}. \] Numerically, \[ u=\frac{580}{(6.91\times10^{-3})(0.80\times10^{-3})} =1.0492\times10^8\ \mathrm{m\,s^{-1}}. \]

The velocity-selection condition becomes \[ v(\phi)=\frac{u}{\sin\phi}. \]

As \(\phi\) decreases from \(90^\circ\), the selected speed increases. The smallest observed angle, \(23^\circ\), therefore corresponds to the maximum speed: \[ v_{\max} =\frac{1.0492\times10^8}{\sin23^\circ} =2.6852\times10^8\ \mathrm{m\,s^{-1}} =0.8951c. \]

The “End of trace” entry specifies this angular cutoff; it supplies no numerical measurement of \(y\).

Relativistic kinetic energy is required: \[ \gamma_{\max} =\frac{1}{\sqrt{1-v_{\max}^2/c^2}} =2.2426, \] \[ K_{\max}=(\gamma_{\max}-1)m_0c^2. \]

Therefore \[ K_{\max} =(2.2426-1)(9.11\times10^{-31})(3.00\times10^8)^2 =1.0188\times10^{-13}\ \mathrm J. \]

Using the standard unit conversion \(1\ \mathrm{eV}=1.602176634\times10^{-19}\ \mathrm J\), \[ K_{\max}=6.36\times10^5\ \mathrm{eV}. \]

To sensible precision for the endpoint angle, \[ \boxed{K_{\max}\simeq6.4\times10^5\ \mathrm{eV} =0.64\ \mathrm{MeV}}. \]

(d)

Outside the plates, we neglect the electric fringe field and retain the leading magnetic deflection. The electron’s horizontal motion is then approximately uniform, and its flight time from the plate edge to the film is \[ \tau\simeq\frac{s}{v}. \]

The magnetic field does no work, so the electron’s speed and Lorentz factor remain constant. To leading order, the vertical magnetic force has magnitude \[ F_z=eB_0v\sin\phi, \] giving vertical acceleration \[ a_z=\frac{eB_0v\sin\phi}{\gamma m_0}. \]

The displacement of either curved trace from the central trace is consequently \[ \delta\simeq\frac12a_z\tau^2 =\frac{eB_0s^2\sin\phi}{2\gamma m_0v}. \]

The two field settings produce opposite displacements. Their separation is therefore \[ y=2\delta \simeq\frac{eB_0s^2\sin\phi}{\gamma m_0v}. \]

Substituting \[ v=\frac{V_0}{B_0t\sin\phi}, \qquad \frac1\gamma =\sqrt{1-\left(\frac{V_0}{B_0tc\sin\phi}\right)^2}, \] we obtain \[ y\simeq \frac{e}{m_0}\frac{B_0^2ts^2}{V_0} \sin^2\phi \sqrt{1-\left(\frac{V_0}{B_0tc\sin\phi}\right)^2}. \]

Since \(0<\phi<180^\circ\), this can be written \[ \boxed{ y\simeq \frac{e}{m_0}\frac{B_0^2ts^2}{V_0} \sin\phi\sqrt{\sin^2\phi- \left(\frac{V_0}{B_0tc}\right)^2} }. \]

Thus choose the graph axes as \[ \boxed{ X=\sin\phi\sqrt{\sin^2\phi- \left(\frac{V_0}{B_0tc}\right)^2}, \qquad Y=y. } \]

The horizontal quantity is dimensionless. Here \[ \frac{V_0}{B_0tc}=0.349735. \]

The points to plot are:

\(\phi\) Horizontal coordinate \(X\) Vertical coordinate \(Y\) / mm
\(90^\circ\) 0.93685 17.4
\(60^\circ\) 0.68612 12.7
\(50^\circ\) 0.52210 9.7
\(40^\circ\) 0.34667 6.4
\(30^\circ\) 0.17867 3.3

The graph is an almost straight line through the origin. Suitable axis ranges are \(0\leq X\leq1\) and \(0\leq Y\leq20\ \mathrm{mm}\). A fit through the origin gives \[ Y\simeq(18.55\ \mathrm{mm})X. \]

Writing the slope in metres, \[ S=1.855\times10^{-2}\ \mathrm m, \] the theoretical relation gives \[ S=\frac{e}{m_0}\frac{B_0^2ts^2}{V_0}. \] Hence \[ \frac{e}{m_0} =\frac{SV_0}{B_0^2ts^2} \simeq \frac{(1.855\times10^{-2})(580)} {(6.91\times10^{-3})^2(0.80\times10^{-3})(41.0\times10^{-3})^2}. \]

Therefore the value inferred from the observations is \[ \boxed{\frac{e}{m_0}\simeq1.68\times10^{11}\ \mathrm{C\,kg^{-1}}}. \]

This derivation retains the leading deflection. At the largest measured separation, \(\delta/s=8.7/41.0\simeq0.21\); finite-curvature corrections can therefore contribute effects of a few percent.

The solution was checked against the official solution.