Two forces on one charge
When an electric field and a magnetic field act together, a charge feels both forces at once. The total is the full Lorentz force, \(\vec F = q(\vec E + \vec v\times\vec B)\). The electric part was the subject of the chapter Electric Charge, Coulomb's Law and the Electric Field. The two parts differ in one important way: \(q\vec E\) is the same at every speed, while \(q\,\vec v\times\vec B\) grows with the speed.
Now arrange the two fields at right angles, as crossed fields. Two parallel plates give an electric field \(\vec E\) down the page, the upper plate being positive. A magnet gives a field \(\vec B\) into the page between the same plates. A proton enters moving to the right. The electric force \(qE\) pushes it down. The magnetic force, by the right-hand rule, pushes it up with \(qvB\).
\[ qE = qvB \]
Why. the proton goes straight only if the two opposite forces are equal in size.
\[ E = vB \]
Why. the charge \(q\) cancels, so the result holds for any charge, positive or negative.
\[ v = \dfrac{E}{B} \]
Read it. solve for the speed; the mass never entered either.
A charge at any other speed is pushed off the straight line. A faster charge feels a magnetic force larger than \(qE\), so it bends the magnetic way. A slower charge bends the electric way. A narrow exit slit then lets through only charges moving at \(E/B\). The device is a velocity selector. For a negative charge both forces reverse together, so it selects the same speed.
Set the speed to \(6\times10^5\,\text{m/s}\). The magnetic force is \(1.2\) times the electric force, so the proton rises and misses the slit by about \(5\,\text{mm}\). At \(4.5\times10^5\,\text{m/s}\) it sags instead. Only at \(5\times10^5\,\text{m/s}\) does it fly straight through. The selected speed is \(E/B = 10^5/0.2 = 5\times10^5\,\text{m/s}\), and it does not depend on the proton's charge or mass.
A velocity selector has \(E = 2\times10^4\,\text{V/m}\) and \(B = 0.01\,\text{T}\). Find the speed it selects, in units of \(10^6\,\text{m/s}\).
In the selector of the last question, an ion enters at \(3\times10^6\,\text{m/s}\). What happens to it?
Thomson weighs the electron
In the 1890s nobody knew what cathode rays were. They streamed from the negative electrode of a tube with most of its air pumped out. J. J. Thomson measured the ratio of their charge to their mass, the specific charge \(e/m\). He used two steps, each giving one equation for the speed.
First, the particles are accelerated from rest through a potential difference \(V\) between the cathode and the anode. The field does work \(eV\) on each one, which becomes kinetic energy: \(\tfrac12 mv^2 = eV\). The chapter Electric Potential, Energy and the Dipole gives this energy. Second, the beam passes between plates carrying crossed fields. Thomson adjusted \(E\) and \(B\) until the glowing spot on the screen did not move. Then \(v = E/B\).
\[ \tfrac12 mv^2 = eV \]
Why. the work done by the accelerating field becomes kinetic energy.
\[ v = \dfrac{E}{B} \]
Why. the undeflected beam balances in the crossed fields.
\[ \tfrac12 m\,\dfrac{E^2}{B^2} = eV \]
Why. put the second speed into the first equation.
\[ \dfrac{e}{m} = \dfrac{E^2}{2VB^2} \]
Read it. rearrange; everything on the right is set or measured on the apparatus.
In a tube like Thomson's, electrons are accelerated through \(400\,\text{V}\). They pass undeflected through crossed fields of \(E = 1.2\times10^4\,\text{V/m}\) and \(B = 1.0\,\text{mT}\). Find their speed and \(e/m\).
The balance gives the speed; the accelerating voltage links the speed to \(e/m\).
\(v = E/B\) and \(e/m = E^2/2VB^2\).
\(v = 1.2\times10^4/10^{-3} = 1.2\times10^7\,\text{m/s}\). \(e/m = (1.2\times10^4)^2/(2\times400\times10^{-6}) = 1.44\times10^8/8\times10^{-4} = 1.8\times10^{11}\,\text{C/kg}\).
The measurement gives \(1.8\times10^{11}\,\text{C/kg}\), within \(3\,\%\) of the modern value \(1.76\times10^{11}\,\text{C/kg}\).
At the Cavendish Laboratory in Cambridge, Thomson measured \(e/m\) for cathode rays in 1897. He changed the metal of the cathode and the gas left in the tube, and the value stayed the same. It was also more than a thousand times the value for a hydrogen ion, the lightest charged thing then known. Thomson concluded that cathode rays were tiny charged particles present in all matter. They came to be called electrons, the first particles found inside the atom.
The measurement gives \(e/m\), not \(e\) and \(m\) separately, because the two appear only as a ratio. Robert Millikan measured \(e\) itself with charged oil drops about ten years later. Together the two results gave the electron's mass.
Thomson found the same value of \(e/m\) whatever metal he used for the cathode. Why was this the decisive result?
If cathode rays were made of charged atoms of the cathode metal, their \(e/m\) would change from one metal to another, because atoms of different elements have different masses. A single value for every metal and every gas meant the particles were the same in all of them. So they had to be a common part of all matter, not a property of one material. That is why Thomson could claim a new particle, the electron.
Weighing ions: the mass spectrometer
A mass spectrometer sorts ions by mass. Ions first pass a velocity selector, so all of them leave it at the same speed \(v = E/B\). They then enter a second region with only a magnetic field \(B_0\). Each ion turns on a half circle and lands on a detector plate, a distance \(2r\) from the slit. From \(r = mv/qB_0\) and \(v = E/B\), the landing point measures the ratio \(m/q\).
The selector is what makes the method work. Without it, a light fast ion and a heavy slow ion could land at the same spot. With every ion at the same speed and charge, the radius is simply proportional to the mass. Each isotope of an element has the same charge and chemistry as the others but a different mass. So the heavier isotope lands farther out.
The radii in Fig 4.1 come from \(r = mv/qB_0\). For neon-20, \(m = 20\,\text{u} = 3.32\times10^{-26}\,\text{kg}\), so \(r = mv/qB_0 = 0.415\,\text{m}\). Neon-22 gives \(0.457\,\text{m}\), ten per cent more. The landing points are twice the radii, \(83.0\,\text{cm}\) and \(91.3\,\text{cm}\) from the slit, so they are \(8.3\,\text{cm}\) apart. In 1913 Thomson found this same pair of neon masses, the first isotopes known for an element that is not radioactive.
In the spectrometer of Fig 4.1, a singly charged ion of neon-21 is also present. How far from the slit does it land, in centimetres? Take \(1\,\text{u} = 1.66\times10^{-27}\,\text{kg}\).
- Thinking the velocity selector picks a particular charge or mass.\(qE = qvB\) gives \(v = E/B\). The charge cancels and the mass never enters.
- Putting the landing point at \(r\) from the slit.The ion turns a half circle, so it lands a full diameter, \(2r\), from the slit.
- Hoping Thomson's method gives \(e\) or \(m\) alone.Both measurements contain only \(e/m\). A separate measurement of \(e\) is needed to find \(m\).