moving-charges Archives

Moving charge; Cyclotron

2021-11-192023-07-13

My notes of the excellent lecture 13 by “Walter Lewin. 8.02 Electricity and Magnetism. Spring 2002. The material is neither affiliated with nor endorsed by MIT, https://youtube.com. License: Creative Commons BY-NC-SA.”

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 “Potentialunterschied”. 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

Proton accelerator
	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	$7,000\,\rm{GeV}$	$\approx\rm c$	$5.5$	$4.2\,\rm{km}$

Electron accelerator
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.

Magnets; charges in motion and Lorenz fore

2021-11-162023-07-13

My notes of the excellent lecture 11 by “Walter Lewin. 8.02 Electricity and Magnetism. Spring 2002. The material is neither affiliated with nor endorsed by MIT, https://youtube.com. License: Creative Commons BY-NC-SA.”

Magnetic fields

In the 5^th century B.C., the Greeks discovered rocks that attract bits of iron. They are named after the area Magnesia where they are plentiful. The rocks contain iron oxide, which we call magnetite.

Magnetites have two places of maximum attraction, the poles. Let’s call them A and B.

A and A repel each other,
B and B repel each other, but
A and B attract each other

There is a huge difference between electricity and magnetism

Electric monopoles exist. If you have a plus charge or minus, that is an electric monopole. If you have a plus and minus of equal strength, you have an electric dipole.
Magnetic poles come in pairs. Magnetic monopoles do not exist, as far as we know. You always have a magnetic dipole.

Poles

The earth is a giant magnet. By convention the part of a magnet that points to the North Canada is called the north pole.

Since A and B attract, that means that North Canada is the magnetic South pole of the earth.

Discovery

In 1819, the Danish physicist Hans Christian Ørsted, discovered that a magnetic needle responds to an electric current in a wire. This linked magnetism with electricity. In this important discovery, Ørsted concluded that the current in the wire produces a magnetic field, and that the magnetic needle moves in response to that.

This discovery caused an explosion of activity in the 19^th century, by André-Marie Ampère, Michael Faraday and Joseph Henry. It culminated in the brilliant work of Scottish theoretician James Clerk Maxwell. Maxwell composed a unified field theory which connects electricity with magnetism. The heart of this course. You will see all “vier” (4) equation at the end of this course.

Coulomb’s law already described the force between electric charges, at rest or in motion. But, these physicists found that the magnetic force is simply an extra electric force acting between moving electric charges.

Moving charges

Remember:

A potential difference generates an electric field along and inside the conductor, that makes the free electrons move. We call this move of charge: current.

Notation convention

Let the wire with a electric current be perpendicular to your screen, so the current goes into the screen.

We put ⊗ to indicate the rear of an arrow. If it would come out of the screen we would the arrow point ⊙.

Current exerts a force on magnets

If we put magnetized needles in the vicinity of the wire with an electric current, they align to form a circle.

This is how we define the magnetic field and its direction, $\vec B$

When the current goes into the screen ⊗, by convention, the magnetic field is in the clockwise direction. To remember that, you can use the righthand corkscrew rule: if you turn it clockwise it goes into the screen.

Magnet exerts a force on wire

Action equals reaction. If the current exerts a force on magnet, then the magnet must exert a force on the wire.

It is an experimental fact that the force on the wire $\vec F$ is always the vectorial cross-product $$ \shaded{ \hat F = \hat I \times \hat B } \nonumber $$

Force between two wires

If I run a current through two wires, it will product a magnetic field. Current $\vec I_1$ will produce $\vec B_1$.

A current $\vec I_2$, based on the righthand rule for the cross-product, $\hat F = \hat I \times \hat B$, the force will be up.

Since action equals minus reaction, the top wire will come down.

Lorenz force

With electricity we defined the strength of an electric field $\vec E$ as the force $\vec F$ on an electric charge $q$

$$ \vec F_{el} = q\,\vec E \nonumber $$

It would be nice if we could define the strength of the magnetic force $\vec F_B$ as a magnetic charge $q_B$ times the magnetic field $\vec B$ $$ \bcancel{ \vec F_B = q_B\,\vec B } \nonumber $$

But there are no magnetic monopoles $q_B$

Force on a charge

Instead, the magnetic field is defined based on an electric charge $q$ moving with velocity $\vec v$ through a magnetic field $\vec B$.

It is an experimental fact, that the force $\vec F_B$ is always perpendicular to $\vec v$. The magnitude of $\vec F_B$ is proportional to the magnitude of $\vec v$, and proportional to the charge $q$ $$ \begin{align*} \vec F_B &\perp \vec v \\ F_B &\propto v \\ F_B &\propto q \end{align*} $$

We define the magnetic field strength as $$ \shaded{ \vec F_B = q\,\left(\vec v \times \vec B\right) } \tag{Lorenz force} $$

This is called the Lorenz force, after the Dutch physicist Hendrik Antoon Lorentz.

The unit is $\frac{N\,s}{C\,m}$ that we call Tesla, $\rm T$. We occasionally use the unit Gauss $\rm G=10^{-4}\,\rm T$.

If you have both electric and magnetic fields, the total force on a particle $\vec F$ $$ \shaded{ \vec F = q\,\left(\vec E + \vec v\times\vec B\right) } \tag{Lorenz force} $$ This is also called the Lorenz force.

An electric field can do work on a charge $$ W = q\,\Delta V \nonumber $$ It can change the kinetic energy of the charge.

Magnetic fields can never do work on a moving charge, because the force $F_B$ is always perpendicular to the velocity $\vec v$. You can change the direction, but you can’t change the kinetic energy.

Force on a whole wire

Calculate the force on a wire with current $I$ through magnetic field $\vec B$. In the wire is charge $dq$ with drift velocity $\vec v_d$

Thought experiments

If the current is zero, at room temperature, the electrons have a huge speed 3 million meters per second, but they are in all chaotic directions (random thermal motion). So on each individual charge there will be a force, but they average out to zero.
It is not until I run a current, that they charges are going to walk through with a very slow drift velocity. Now, the net force will not be zero.

The angle $\theta$ will come in later, because the force is the cross-product between $\vec v$ and $\vec B$.

Side note: in reality negative charges (electrons) move in the opposite direction as the current. However, a negative charge going opposite to the current, is mathematically the same as a positive charge going with the current.

On charge $dq$ there is a force $d\vec F_B$ $$ \begin{align*} d\vec F_B &= dq \, \left(\vec v_d \times \vec B\right), & \left(I = \frac{dq}{dt}\right) \\ &= I\,dt \, \left(\vec v_d \times \vec B\right) \end{align*} $$

The drift velocity $\vec v_d$ is the derivative of the distance $d\vec l$ in time,

With $v_d = \dfrac{d\vec l}{dt}$ $\Longrightarrow$ $d\vec l = \vec v\, dt$ $$ \shaded{ d\vec F_B = I\, \left( d\vec l \times \vec B \right) } \tag{Force} \label{eq:forcedq} $$

This is the force of a small segment of the wire, with length $d\vec l$. $I$ is the current through the wire, and $\vec B$ is the local magnetic field at location $d\vec l$.

If you want to know the entire force on the wire, you have to take the integral along the whole wire. At every potion $d\vec l$, you have to determine what $\vec B$ is. That will give you a force vector that you have to add those vectors vectorially. Not easy, but that’s the idea. $$ \begin{align*} \int_\rm{wire} d\vec F_B &= \int_\rm{wire} I\, \left( d\vec l \times \vec B \right) \end{align*} $$

Example

Simplifying the geometry so we can solve the integral. Assume the magnetic field was constant over the portion of the wire.

With $\vec B\perp d\vec l$, so $\sin\theta=1$. And $\vec B$ is constant $(B)$. $$ \begin{align*} F_B &= \int_\rm{wire} I\, \left( d\vec l \times \vec B \right) \\ &= I\, \left( \int_\rm{wire} d\vec l \right) \times B \end{align*} $$

The current $\vec I$ and magnetic field $\vec B$ are constant, so only $d\vec l$ stays in the integral $$ \shaded{ F_B = I\, l \, B } \tag{Force on wire} \label{eq:forcewire} $$

Filling in the values $$ \begin{align*} F_B &= I\, \left( \int_\rm{wire} d\vec l \right) \times B \\ &= I\, l \, B \\ &= (300)(0.1)(0.2) = 6\,\rm N \end{align*} $$

DC Motors

Current meter

A constant current loop in a constant magnetic field. The force $\vec F$ is in the direction $\vec I\times\vec B$

36:45

If the wire has length $a$, then $B$ with $\vec B\perp \vec I$, the force is what we just derived in equation $\eqref{eq:forcewire}$ $$ \begin{align*} F &= I\,a\,B \end{align*} $$

On the front and rear sections, the force is zero, because the magnetic field and current are parallel, so their cross-product is zero.

As a result of the forces, there is a torque that wants to rotate it in the counter clockwise direction. The the distance between the forces $b$, the magnitude of the torque at this moment in time $$ \tau = 2F\,\frac{1}{2}b = I\,a\,b\,B \nonumber $$

As it rotates, the forces come closer, so the torque will be come less. There will come a time, 90° later, that the torque is zero.

At that time, the whole loop is perpendicular to the magnetic field. The forces follow from the cross-product rule

At that point in time, there is again no net force on the system, but now there is no torque either. At it reaches 90°, it has enough inertia, so that it rotates a little further. Now the torque will reverse, causing it to rotate back.

Commutator

How do you get over this torque reversal and the wires to $A$ and $D$ intertwining?

The solution is slippery contacts, called brushes.

This not only keeps the wires from tangling up, it also reverses the potential. It it rotates 180° then A will be on the minus side of the potential. Now every half a rotation, the current will reverse direction. We call that a commutator.

Now, the torque reversal will not occur. The torque will always want to rotate the loop in exactly the same direction.

	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
\(15\,\rm{eV}\)	\(2.3\times 10^6\)	\(0.5\times10^{-4}\)	\(26\,\rm{cm}\)
\(100\,\rm{keV}\)	\(1.6\times 10^8\)	\(1\)	\(1.1\,\rm{mm}\)
\(500\,\rm{keV}\)	\(2.6\times10^8\)	\(1\)	\(2.9\,\rm{mm}\)

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