\(\)

## Path of a moving charge in \(\vec B\)

Remember: Lorenz force \(\vec F\) depends on the velocity of the charge and the magnetic field that it experiences

$$ \vec F = q\,\left(\vec v\times\vec B\right) \nonumber $$

If we have the following charge \(q\) and velocity \(\vec v\), it will experience \(\vec F\)

This causes the charged particle to go around in a perfect circle. The Lorenz force can’t change the speed (can’t change the kinetic energy), because \(\vec F\perp \vec v\), but it **can change the direction of the velocity**.

The radius \(R\) of the circle follows from the force. With \(\vec F\perp \vec v\), the \(\sin\) of the angle between them is \(1\). That is now the centripetal force \(\frac{mv^2}{R}\), where \(m\) is the mass of the particle $$ F = q\,v\,B = \frac{mv^2}{R} \nonumber $$

So the radius $$ \shaded{ R = \frac{m\,v}{q\,B} } \label{eq:R} $$

Here \(mv\) is the momentum.

### Numerical example

\(U\)Instead of \(\Delta V\), we often use the European symbol for voltage difference: \(U\). The letter ‘u’ stands for “Potential__u__nterschied”. Do not confuse it with potential energy, that also uses \(U\).

Assume a proton with kinetic energy \(1\,\rm{MeV}\) \(=1.6\times10^{-13}\,\rm J\) in a constant magnetic field \(\vec B\) $$ \begin{align*} q\,\Delta V &= 1.6\times10^{-13}\,\rm J \\ m_p &= 1.7\times 10^{-27}\,\rm{kg} \\ q_p &= 1.6\times 10^{-19}\,\rm{C} \end{align*} $$

The speed of the proton $$ \begin{align} q\,\Delta V &= \frac{1}{2}m\,v^2 \nonumber \\ \Rightarrow v &= \sqrt{\frac{2\,q\Delta V}{m_p}} \nonumber \\ &\approx \sqrt{\frac{2\,(1.6\times10^{-13})}{1.7\times 10^{-27}}} \nonumber \\ &\approx 1.4\times 10^7\,\rm{m/s} \approx 0.05\,\rm c \label{eq:onemkvproton} \end{align} $$

That is 5% of the speed of light, comfortably low, so we don’t have to make any relativistic corrections.

If this proton now enters the magnetic field \(B=1\,\rm T\). Based on equation \(\eqref{eq:R}\), it will circle with radius \(R\) $$ \begin{align*} R &= \frac{mv}{qB} \\ &= \frac{(1.7\times10^{-27})(1.4\times 10^7)}{(1.6\times 10^{-19})(1)} \\ &\approx 1.5\,\rm m \end{align*} $$

### Eliminate \(\text{ }\vec v\)

In equation \(\eqref{eq:R}\), the velocity \(v\) is commonly replaced by the potential difference \(\Delta V\) over which we accelerate these particles.

Find \(v=f(\Delta V)\) $$ \begin{align*} q\,\Delta V &= \frac{1}{2}m\,v^2 \\ \Rightarrow v &= \sqrt{\frac{2\,q\,\Delta V}{m}} \end{align*} $$

Substituting this in equation \(\eqref{eq:R}\) $$ \begin{align*} R &= \frac{m\,\sqrt{\frac{2\,q\,\Delta V}{m}}}{qB} \end{align*} $$

That simplifies to $$ \shaded{ R = \sqrt{\frac{2\,m\,\Delta V}{q\,B^2}} } \tag{Radius} $$

### Approaching \(\text{ c}\)

When the speed approaches the speed of light, we have to apply special relativity. That is not part of this course, but we will briefly touch upon that

Things go sour, when we have a \(500\,\rm{keV}\) electron, then $$ \begin{align*} \frac{1}{2}m\,v^2 &= q\,\Delta V \\ \Rightarrow v &= \sqrt{\frac{2\,q_e\,\Delta V}{m_e}} \\ &= \sqrt{\frac{2\,(1.6\times10^{-19})\,(500\times10^3)}{9.1\times 10^{-31}}} \\ &\approx 4.2\times10^8\,\rm{m/s} \gt \rm c \end{align*} $$

This is larger than the speed of light, so clearly not possible. If you make relativistic corrections, you will find the actual speed. Just to show you, we’ll make these corrections below.

#### Relativistic correction

Start with same kinetic energy \(q\,\Delta V\), but it is no longer \(\frac{1}{2}m\,v^2\) $$ q\,\Delta V = (\gamma – 1)\,m\,c^2 \nonumber $$

Here the **Lorenz factor** \(\gamma\) is defined as
$$
\gamma = \frac{1}{\sqrt{1 – \frac{v^2}{c^2}}}
\tag{Lorenz factor}
$$

The Lorenz factor follows as $$ \begin{align*} q\,\Delta V &= (\gamma – 1)\,m\,c^2 \\ \Rightarrow \gamma &= \frac{q\,\Delta V}{m\,c^2} + 1 \\ &\approx \frac{(500\times 10^3)(1.6\times 10^{-19})}{(9.1\times10^{-31})\,(300\times10^6)^2} + 1 \\ &\approx 1.977 \end{align*} $$

The speed \(v\) $$ \begin{align*} \sqrt{1 – \frac{v^2}{c^2}} &= \frac{1}{\gamma} \Rightarrow \frac{v^2}{c^2} \\ \Rightarrow v &= c\,\sqrt{1 – \frac{1}{\gamma^2}} \\ \Rightarrow v &= 300\times10^6 \sqrt{1 – \frac{1}{1.977^2}} \\ &\approx 2.6\times 10^8\,\rm{m/s} \end{align*} $$

The radius \(R\) in a \(1\,\rm T\) magnetic field $$ \begin{align} R &= \gamma\,\frac{m\,v}{q\,B} = \sqrt{\frac{(\gamma+1)\,m\,\Delta V}{q\,B^2}} \label{eq:corrR} \\ &= 1.977\,\frac{(9.1\times10^{-31})\,(2.6\times 10^8)}{(1.6\times10^{-19})\,1} \nonumber \\ &\approx 2.9\,\rm{mm} \nonumber \end{align} $$

### Some examples

KE | v [m/s] | B [T] | R | |
---|---|---|---|---|

\(1\,\rm{MeV}\) | \(1.4\times 10^7\) | \(1\) | \(0.15\,\rm{m}\) | |

\(50\,\rm{MeV}\) | \(9.4\times 10^7\) | \(1\) | \(1.04\,\rm{m}\) | |

Fermilab | \(500\,\rm{GeV}\) | \(2.99999\times10^8\) | \(1.5\) | \(1.1\,\rm{km}\) |

LHC | \(7,000\,\rm{GeV}\) | \(\approx\rm c\) | \(5.5\) | \(4.2\,\rm{km}\) |

KE | v [m/s] | B [T] | R | |
---|---|---|---|---|

\(15\,\rm{eV}\) | \(2.3\times 10^6\) | \(0.5\times10^{-4}\) | \(26\,\rm{cm}\) | |

\(100\,\rm{keV}\) | \(1.6\times 10^8\) | \(1\) | \(1.1\,\rm{mm}\) | |

\(500\,\rm{keV}\) | \(2.6\times10^8\) | \(1\) | \(2.9\,\rm{mm}\) |

### Practical use

In 1945, the Americans needed Uranium \({}^{235}\rm U\) to build an atomic bomb. This has 143 neutrons, instead of the more common 146. To separate them, Ernest Lawrence of Berkeley built a mass spectrometers that could separate \({}^{235}\rm U\) and \({}^{238}\rm U\).

In a mass spectrometer, the uranium is heated so that it ionizes. Let’s assume it looses one electron, so it’s positively charged with one unit charge. It then accelerates it over a potential difference, so they get a certain speed.

In a constant magnetic field, the charged particles go around a circle and hit a collector. The radius of the circle is proportional with \(\sqrt{m}\). The mass of 238 is 1.2% larger than the mass of 235. So they land is a collector slightly separated from the other one.

## Accelerating protons

In 19939, the UC Berkeley physicist, Ernest Laurence received a Nobel Prize for Physics. He invented the cyclotron that accelerates protons to \(730\,\rm{MeV}\), almost the speed of light.

### Cyclotron

In the early days accelerating was done in a cyclotron. It consists of two conducting chambers \(D\) in vacuum. Seen from the top and side:

Suppose we release a \(1\,\rm{MeV}\) proton. That comes out a speed of \(1.4\times10^7\,\rm{m/s}\) \(\eqref{eq:onemkvproton}\). In a \(1\,\rm{T}\) field, based on the earlier table, the radius is going to be \(15\,\rm{cm}\).

The proton start to make a circle. But when it gets halfway, a potential difference is introduced between these two \(D\)’s. The high potential on the right \(D\), and the low potential on the left \(D\). In the gap between the \(D\)’s, you get an electric field \(\vec E\)

The field accelerates the proton. If the potential difference is \(20\,\rm{kV}\), the proton will gain \(20\,\rm{keV}\) kinetic energy. When it crossed the gap, it has \(1.02\,\rm{MeV}\). If \(V\) is 2% higher, then based on equation \(\eqref{eq:corrR}\), the radius is 1% higher than \(15\,\rm{cm}\). It comes out a little higher.

Once, it gets to the top, the potential difference is reversed, so the \(\vec E\) points to the right, and it is again accelerated by \(20\,\rm{keV}\).

Very gradually, it spirals out the largest radius. So for every rotation it gains \(40\,\rm{keV}\). The electric fields are doing the work, they accelerate the particles. The magnetic field changes the direction, but they can’t do work on the particles. It confines the particles.

If we go 1225 full rotations, where each time the kinetic energy increases by \(40\,\rm{keV}\). Multiplying these you find that it gained \(49\,\rm{MeV}\). Plus the initial \(1\,\rm{MeV}\), it now has \(50\,\rm{MeV}\). Referring to the the earlier table, the radius is one meter.

If we don’t have to make relativistic corrections, the time to go around is independent on the proton’s speed. With the radius proportional to \(v\), so the time is independent of \(v\) $$ T = \frac{2\pi R}{\rm{speed}} = \frac{2\pi\,m}{q\,B} \nonumber $$

The time to go around once, is only \(66\,\rm{ns}\). To go around 1,225 times will take only \(80\,\rm{msec}\). That means the switching frequency becomes about \(30\,\rm{MHz}\).

Because of relativistic correction, the time to go around once, becomes $$ T = \frac{2\pi\,m}{q\,B}\,\gamma = 6.6\times 10^{-8}\,\rm{sec} \nonumber $$

If you go to very high energies, that time is not constant for a full rotation. So you have to adjust the switching frequency. We call those instruments *synchrotrons* or *synchrocyclotrons*.

### Modern accelerators

Modern accelerators have constant radii. They are rings. The only way to keep the particles in the ring when they have a low energy, and when they have a high energy, is by gradually increasing the magnetic field. By making the magnetic field go up, you can keep the protons in that ring.

#### Fermilab

Fermilab near Chicago. One of the modern accelerators, occasionally called colliders. A diameter of \(2.2\,\rm{km}\), accelerates up to \(1000\,\rm{GeV} = 10^{12}\,\rm{eV}\). The beams of high energy protons are made collide with other nuclei to undercover the inner workings of nuclear physics.

#### Large Hadron collider

The tunnel of the largest ring in the work, at CERN in Geneva, has a radius of \(4.3\,\rm{km}\). Using superconducting magnets, they can go to about \(5\,\rm T\). The Large Hadron collider accelerates protons to the unprecedented energy of \(6500\,\rm{GeV} = 6.5\times 10^{12}\,\rm{eV}\).

##### Higgs boson

In July 2012, the LHC made one of the most important discoveries in particle physics. They proved the existence of the Higgs boson, which was hypothesized about 45 years earlier by Peter Higgs.

The Higgs boson is part of what’s called the Standard Model of particle physics. A set of rules that lays out our understanding of the fundamental building blocks of the universe. The Higgs boson has a mass of about \(125\,\rm{GeV/c^2}\), about \(133\) proton masses.

In 2013, Peter Higgs and François Englert were awarded the Nobel Prize in physics for their work in identifying and discovering the Higgs boson.