Thomson could measure the deflection of cathode-ray particles for various strengths of the magnetic field, but magnetic deflection depends both on the particle’s charge-to-mass ratio $q\text{/}m$ and on its speed. Measuring the charge-to-mass ratio, and thus learning something about the particles themselves, requires some means of determining their speed. To do so, Thomson devised the experiment for which he is most remembered.

Thomson built a tube containing the parallel-plate electrodes visible in the photo in FIGURE 37.12a. He then placed the tube in a magnetic field. FIGURE 37.12b shows that the electric and magnetic fields were perpendicular to each other, thus creating what came to be known as a crossed-field experiment.

FIGURE 37.12 Thomson’s crossed-field experiment to measure the velocity of cathode rays. The photograph shows his original tube.

Full Alternative Text

The magnetic field, which is perpendicular to the particle’s velocity $\vec{v}$, exerts a magnetic force on the charged particle of magnitude

The magnetic field alone would cause a negatively charged particle to move along an upward circular arc. As you learned in 《 SECTION 29.7, the radius of the arc is

It is a straightforward geometry problem to determine the radius of curvature r from the measured deflection of the green spot.

Thomson’s new idea was to create an electric field between the parallel-plate electrodes that would exert a downward force on the negative charges, pushing them back toward the center of the tube. The magnitude of the electric force on each particle is

Thomson adjusted the electric field strength until the cathode-ray beam, in the presence of both electric and magnetic fields, had no deflection and was seen exactly in the center of the tube.

Zero deflection occurs when the magnetic and electric forces exactly balance each other, as FIGURE 37.12c shows. The force vectors point in opposite directions, and their magnitudes are equal when

Notice that the charge q cancels. Once E and B are set, a charged particle can pass undeflected through the crossed fields only if its speed is

By balancing the magnetic force against the electric force, Thomson could determine the speed of the charged-particle beam. Once he knew v, he could use Equation 37.6 to find the charge-to-mass ratio:

Thomson found that the charge-to-mass ratio of cathode rays is $q\text{/}m\approx 1\times {10}^{11}\text{ C/kg}$. This seems not terribly accurate in comparison to the modern value of $1.76\times {10}^{11}\text{C}\text{/}\text{kg}$, but keep in mind both the experimental limitations of his day and the fact that, prior to his work, no one had any idea of the charge-to-mass ratio.

EXAMPLE 37.2 A crossed-field experiment

An electron is fired between two parallel-plate electrodes that are 5.0 mm apart and 3.0 cm long. A potential difference $\text{Δ}V$ between the electrodes establishes an electric field between them. A 3.0-cm-wide, 1.0 mT magnetic field overlaps the electrodes and is perpendicular to the electric field. When $\text{Δ}V=0\text{ V}$, the electron is deflected by 2.0 mm as it passes between the plates. What value of $\text{Δ}V$ will allow the electron to pass through the plates without deflection?

MODEL Assume that the fields between the electrodes are uniform and that they are zero outside the electrodes.

VISUALIZE FIGURE 37.13 shows an electron passing through the magnetic field between the plates when $\text{Δ}V=0\text{V}$. The curvature has been exaggerated to make the geometry clear.

FIGURE 37.13 The electron’s trajectory in Example 37.2.

FIGURE 37.13 Full Alternative Text

SOLVE We can find the needed electric field, and thus $\text{Δ}V$, if we know the electron’s speed. We can find the electron’s speed from the radius of curvature of its circular arc in a magnetic field. Figure 37.13 shows a right triangle with hypotenuse r and width L. We can use the Pythagorean theorem to write

$${(r-\text{Δ}y)}^2+L^2=r^2$$

where $\text{Δ}y$ is the electron’s deflection in the magnetic field. This is easily solved to find the radius of the arc:

$$r={\displaystyle \frac{{\left(\text{Δ}y\right)}^2+L^2}{2\text{Δ}y}}={\displaystyle \frac{(0.0020{\text{m)}}^2+(0.030{\text{m)}}^2}{2(0.0020\text{m})}}=0.226\text{m}$$

The speed of an electron traveling along an arc with this radius is found from Equation 37.6:

$$v={\displaystyle \frac{erB}{m}}=4.0\times {10}^7\text{m}\text{/}\text{s}$$

Thus the electric field allowing the electron to pass through without deflection is

$$E=vB=40\text{,}000\text{V}\text{/}\text{m}$$

The electric field of a parallel-plate capacitor of spacing d is related to the potential difference by $E=\text{Δ}V\text{/}d$, so the necessary potential difference is

$$\text{Δ}V=Ed=(40\text{,}000\text{V}\text{/}\text{m})(0.0050\text{m})=200\text{V}$$

REVIEW A fairly small potential difference is sufficient to counteract the magnetic deflection.