### Electric Force

By Charles deAnne

Most recent revision: February 21, 2020

To know that we know what we know, and to know that we do not know what we do not know, is true knowledge.

This web page is a work in progress. Any suggestions or corrections are welcome and will be credited.
email: charlesdeanne@sbytes.com

#### Summary of results:

The number of quantum electric particles in a coulomb is given by:

$1 coulomb=\sqrt{\frac{10^{-7}newton\cdot second^{2}}{r_{e}m_{e}}}$

Note that the physical dimensions cancel in the above equation, making the coulomb a pure number. It is an arbitrary quantity determined by Ampere's decision to fix the magnetic force per meter between two parallel coulomb-per-second currents separated by one meter to 2X10-7 newton.

$\therefore 1 coulomb=\sqrt{\frac{10^{-7}}{(2.817 940 3e-15 )(9.109 383 7e-31)}}=6.2415091\times 10^{18}$

The electron charge is the reciprocal of the coulomb.

$\therefore e^{2}=\frac{r_{e}m_{e}}{10^{-7}}\sim \left ( coulomb/electron \right )^{2}$

Electron charge, e, also has no physical dimensions.

The electrostatic constant defines the force between point charge concentrations:

$F=\frac{r_{e}m_{e}c^{2}}{r^{2}}\frac{Q_{Y}Q_{Z}}{e^{2}}=\frac{c^{2}\times 10^{-7}}{r^{2}}Q_{Y}Q_{Z}=\frac{k_{e}}{r^{2}}Q_{Y}Q_{Z}$

$\therefore k_{e}=c^{2}\times 10^{-7}\sim \frac{newton\cdot meter^{2}}{coulomb^{2}}$

The equation for the constant electric field between two parallel planes with opposite but equal coulomb surface densities, sigma_q, contains an embarrassing factor of 4PI when expressed as a function of the electrostatic constant.

$E=F_{q}=4\pi r_{e}m_{e}c^{2}\sigma_{q}\left ( \frac{10^{-7}}{r_{e}m_{e}} \right ) =4\pi k_{e}\sigma_{q}$

To simplify the above equation (and eliminate the problematic four-pi factor), the electrostatic constant was re-defined as:

$k_{e}=\frac{1}{4\pi \epsilon _{0}}$

Epsilon_zero, or permittivity of free space, now gives the force per coulomb on a charge located between parallel planes of equal and opposite coulomb densities as:

$E=\frac{\sigma _{q}}{\epsilon _{0}}$

Permeability of free space, mu_zero, was originally defined as:

$\mu _{0}=\frac{1}{\epsilon _{0}c^{2}}$

$\mu _{0}=\frac{4\pi r_{e}m_{e}}{e^{2}}=4\pi \times 10^{-7}newton\cdot second^{2}/coulomb^{2}\sim newton/ampere^{2}$

This also gives the truly mystical definition of light speed:

$c^{2}=\frac{1}{\epsilon _{0}\mu _{0}}$

#### Potential energy between two quantum electric particles

The absolute value of potential energy between two quantum electric particles (electrons or protons), separated by the classical electron radius, is equal to the rest mass energy of the electron. Since potential energy is inversely proportional to the distance of separation, the equation for potential energy between each unique pair of electric particles is:

$u_{p}=\frac{r_{e}m_{e}c^{2} }{r}(\pm 1)(\pm 1)$

The factors, plus or minus one, represent single particles at each end of r, with a plus assigned to protons and a minus assigned to electrons. For point concentrations of Y particles at one end of r and Z particles at the other end, the total number of unique quantum pairs is the product, YZ, giving the total potential energy as:

$u_{p}=\pm \frac{r_{e}m_{e}c^{2} }{r}YZ$

#### Force between point concentrations of quantum particles

A spatial gradient of energy is equivalent to a force directed toward the region of lower energy. Taking the derivative of the above equation with respect to r gives the electric force between two point concentrations of quantum particles:

$f=\pm \frac{r_{e}m_{e}c^{2} }{r^{2}}YZ$

#### Force between a single electron and an infinite straight line of electrons

Imagine a line of uniformly distributed electrons on the 'x' axis, and the subject single electron on the 'y' axis, distance, d, from the origin. Partitioning the 'x' axis into small, equal, incremental distances, dx, the number of electrons in each partition is:

$Z_{x}=\lambda _{Z}dx$
(lambda_Z is the linear density in electrons per unit length)

For each dx at +x, there is a corresponding dx at -x. Since the horizontal components of force from each paired group of electrons are equal and opposite, they cancel each other, leaving the net force always perpendicular to the 'x' axis:

$\Delta F=2\cos \phi \frac{r_{e}m_{e}c^{2}}{L^{2}}\lambda _{Z}dx$

(L is the distance from the electron to the dx located at +x, and phi is the acute angle between the 'y' axis and L.)

$L=\frac{d}{\cos \phi }$

Substituting and re-grouping,

$\Delta F=2\lambda _{Z}r_{e}m_{e}c^{2}\frac{\cos \phi }{d}\frac{\cos \phi dx}{L}$

The differential angle subtended by dx is the component of dx perpendicular to L, divided by L:

$d\phi =\frac{\cos \phi dx}{L}$

$\therefore \Delta F=\frac{2\lambda _{Z}}{d}r_{e}m_{e}c^{2}\cos \phi d\phi$

Integrating the incremental forces while phi goes from zero to PI/2, is equivalent to integrating the same incremental forces as x goes from zero to infinity. Since:

$\int_{0}^{\frac{\pi }{2}}\cos \phi d\phi =1$

The total vertical force on the electron is:

$F=\frac{2\lambda _{Z}}{d}r_{e}m_{e}c^{2}$

#### Force between parallel lines of quantum particles

By considering the above subject electron to be part of a parallel line of uniformly distributed electrons, with lambda_Y being its linear density, a general equation emerges for the attractive or repulsive force per unit length between two parallel linear quantum densities:

$\frac{dF}{dL}=\pm 2\frac{r_{e}}{d}m_{e}c^{2}\lambda _{Y}\lambda _{Z}$
(L here represents length in the same direction as the parallel lines of particles.)

The above equations describe static electric forces which are repulsive if the signs of both quantum lines of particles are the same and attractive if their signs are opposite.

#### Magnetic force between parallel currents

Now consider two long, straight, parallel, electrically neutral conductors.  If the conductors carry currents in which the rates of electron flow are given by $\dot{Y}$ and $\dot{Z}$, respectively, the equivalent electron densities for light-speed currents are given by:

$\lambda_{Y}=\frac{\dot{Y}}{c};\lambda_{Z}=\frac{\dot{Z}}{c}$
(Particles per second divided by meters per second is equal to particles per meter.)

Substituting these equivalent electron densities into the static equation:

$\frac{dF}{dL}=\pm 2 \frac{r_{e}}{d}m_{e}\dot{Y}\dot{Z}$

This equation expresses the quantum origin of magnetic force. If both currents consist of electrons moving in the same direction, the magnetic force is attractive; otherwise it is repulsive.

Equivalent linear electron density is the hypothetical electron density for a light-speed current. It will be demonstrated that these equivalent electron densities in the two conductors are invisible to each other.

Consider two parallel electron currents. The repulsive force between the two equivalent electron densities was an essential component of equilibrium in the static case, absent the currents. The loss of this repulsive force in the presence of the currents can be interpreted as the cause of the observed attraction between the two conductors.

While it is easy to visualize how the loss of repulsion between two parallel light-speed electron densities might lead to an observed attraction, the observed repulsion between opposite electron flows is not quite as intuitive; unless that is, we accept John Wheeler's conjecture reported by Richard Feynman in his Nobel lecture. To paraphrase: "An electron moving in one direction is indistinguishable from a positron moving in the opposite direction." Opposite direction electron currents can then be visualized as one positron current and one electron current, both flowing in the same direction. The loss of their static attraction becomes the source of their observed repulsion.

This is the basic model for magnetic force which Nand tried to explain to John Wheeler in the “The Matterhorn Paradox”. Wheeler had not yet fully embraced his own conjecture--he wasn’t ready for this leap.

#### Determining the number of electrons in a coulomb

In the previous section, the magnetic force per unit length between two parallel currents was derived by integrating small contributions over an infinite length of the parallel currents:

$\frac{dF}{dL}=\pm 2 \frac{r_{e}}{d}m_{e}\dot{Y}\dot{Z}$

$\dot{Y}$and $\dot{Z}$ are the electron flow rates in the two currents, re is the classical electron radius, me is the electron mass, and d is the distance separating the two currents.

In 1820, Andre-Marie Ampere defined his namesake unit of current such that one unit of current in both of two parallel straight conductors, separated by one meter, would produce 2 X 10-7N of force between each meter length of the conductors. While his definition was totally arbitrary, it was obviously chosen to give a simple base-ten value for the magnetic force in newton per meter. His choice of two times a power of ten suggests that Ampere had solved the above integral in a general form, without knowing the product of the classical electron radius and the electron mass. When one meter is substituted for d, and one coulomb per second is substituted for both $\dot{Y}$and $\dot{Z}$, the factors of two cancel, assuring a simple relationship between the coulomb and the unknown (in 1820) constant, reme, whatever its value might be.

$\frac{dF}{dL}=2\frac{r_{e}}{1 meter}m_{e}\frac{coulomb^{2}}{second^{2}}=2\times 10^{-7}newton/meter$

$1 coulomb=\sqrt{\frac{10^{-7}newton\cdot second^{2}}{r_{e}m_{e}}}$
Note that the physical dimensions cancel in the above equation, making the coulomb a pure number.

$\therefore 1 coulomb=\sqrt{\frac{10^{-7}}{(2.817 940 3e-15 )(9.109 383 7e-31)}}=6.2415091\times 10^{18}$

This was seventy-seven years before J.J. Thomson measured the mass/charge ratio of the electron, and eighty-seven years before Robert Millikan is credited with the measurement of the electron charge. If Ampere had known the electron's mass, he undoubtedly would have calculated the number of electrons in a coulomb from the above equation.

As it happened, however, fourteen years after Ampere defined the unit of current, Michael Faraday discovered that the coulomb could be measured experimentally from the electroplating mass-rate caused by one ampere of current. For nearly two hundred years physicists have relied on this empirical definition of the coulomb. There has been little interest in exploring the underlying quantum interactions responsible for its value.

The coulomb is a ubiquitous component of all other electromagnetic constants. In the absence of a theoretical understanding of its value, the coulomb and its reciprocal, the electron charge, appeared to be themselves constants of nature. From the above demonstration it is clear that the coulomb's value is very much dependent on the SI system of physical units as well as being defined with a bias for base-ten numbers. To now consider the coulomb and electron charge to be constants of nature, one must also assume that nature has a preference for force, length, and time to be measured in newton, meter, and second as well as a preference for numbers that are integer powers of ten.

Eggs are sold by the dozen.
Electrons are sold by the coulomb.
Electron charge is the reciprocal of the coulomb.
Therefore, the electron charge is a property of the electron in the same way a twelfth dozen is a property of an egg.

#### The electrostatic constant

Now that we have an explicit definition of the coulomb, it might be informative to revisit the equation for the electric force between point concentrations of quantum particles:

$f=\pm \frac{r_{e}m_{e}c^{2}}{r^{2}}YZ=\pm \frac{k_{e}}{r^{2}}YZ$

By substituting one coulomb each for both Y and Z, and one meter for r, the force between two coulomb-concentrations of quantum particles, separated by one meter is:

$f=\pm \frac{r_{e}m_{e}c^{2}}{1meter^{2}}\left ( \frac{10^{-7}newton\cdot second^{2}}{r_{e}m_{e}}\right)=c^{2}\times 10^{-7}newton$

Since the parenthetical expression is coulomb2, it can be replaced by its reciprocal, (electron charge squared) in the denominator. The electrostatic constant can, therefore, be defined as:

$k_{e}=\frac{r_{e}m_{e}}{e^{2}}c^{2}=c^{2}\times 10^{-7}newton\cdot meter^{2}/coulomb^{2}$

While the electrostatic constant is numerically equal to the speed of light (in meters per second) squared times ten to the minus seven, its physical dimensions are cryptically remote from velocity-squared. (The 10-7 carries hidden dimensions of: newton second2)

Dividing by c2, yields the seemingly improbable, but exact evaluation:

$\frac{r_{e}m_{e}}{e^{2}}=10^{-7}newton\cdot second^{2}/coulomb^{2}$

#### Force between a single electron and a plane of uniformly distributed protons

Dropping a line from the electron, perpendicular to the plane, finds the closest point on the plane. Using that point as the center, consider the plane to be partitioned into concentric circular rings with radius, r, and width, dr. If $\sigma _{Z}$ is the proton surface density in protons per unit area , the number of protons in each ring is:

$Z_{r}=\sigma _{Z}2\pi r dr$

All of the protons in any ring are on a cone, equal distance, L, from the electron:

$L=\frac{r}{\sin \phi }$

From symmetry, it is clear that any force component parallel to the plane will be cancelled by equal proton concentrations on opposite sides of the ring. The total force contribution from any ring is, therefore, only its component perpendicular to the plane:

$F_{r}=r_{e}m_{e}c^{2}\frac{Z_{r}}{L^{2}}\cos \phi =r_{e}m_{e}c^{2}\frac{\sigma _{Z}2\pi rdr}{L^{2}}\cos \phi$

The differential angle subtended by dr is the component of dr perpendicular to L, divided by L:

$d\phi =\frac{dr }{L}\cos \phi$

$\therefore F_{r} =2\pi r_{e}m_{e}c^{2}\sigma _{Z}\frac{r}{L}d\phi =2\pi r_{e}m_{e}c^{2}\sigma _{Z}\sin \phi d\phi$
Since,

$\int_{0}^{\frac{\pi }{2}}\sin \phi d\phi =1$
, the total force on one electron, any distance from an infinite plane of protons is:

$F_{e}=2\pi r_{e}m_{e}c^{2}\sigma _{Z}$

#### Pressure between two parallel planes of uniformly distributed electric particles

If the above single electron happens to be part of a uniformly distributed parallel plane of electrons, the force per area (pressure) between the two planes is:

$F/A=2\pi r_{e}m_{e}c^{2}\sigma _{Z}\sigma _{Y}$

#### Force on a single electron between two planes of opposite charge

Since the electron is repelled by the negative plane, and attracted to the positive plane, the two forces are in the same direction and, therefore, additive.

#### $F_{e}=2\pi r_{e}m_{e}c^{2}\left (\sigma _{Z}+\sigma _{Y} \right )$

If the quantum surface densities are equal and opposite, as is normally the case with a capacitor:

$F_{e}=4\pi r_{e}m_{e}c^{2}\sigma$

#### Permittivity of free space--artifact from an age of ignorance

The above equation gives the force on a single electron, with sigma representing the quantum surface density on both parallel planes. To convert the equation to force per coulomb, it must be multiplied times coulomb squared. (One coulomb to replace the single electron with one unit of charge, the other to express sigma in coulombs per unit area.)

Force per coulomb on a point concentration of charge between two planes with opposite charge, but equal coulomb densities, sigma_q, (aka 'E'):

$E=F_{q}=4\pi r_{e}m_{e}c^{2}\sigma_{q}\left ( \frac{10^{-7}}{r_{e}m_{e}} \right ) =4\pi k_{e}\sigma_{q}$
To simplify the above equation even further (and eliminate the problematic four-pi factor), the electrostatic constant was re-defined as:

$k_{e}=\frac{1}{4\pi \epsilon _{0}}$

Giving the force per coulomb on charges located between parallel planes of equal and opposite coulomb densities as:

$E=\frac{\sigma _{q}}{\epsilon _{0}}$

Epsilon_zero was given the intimidating tittle, "Permittivity of free space", implying that it is a universal property of space. Whether intended or not, this moniker cloaked the fact that its value was artificially contrived. At the same time, it discouraged curiosity about its true origin or physical meaning.

In retrospect, it is understandable that the 4pi factor was troublesome, given the established Gaussian model of electric flux density through a surface giving rise to electric force. The validity of this model seemed obvious from the following observations of force exerted by various distributions of electric particles on point concentrations of electric particles:
* For distance, r, from point concentrations, force varies inversely with the surface area of an imaginary sphere (1/r2).
* For distance, r, from a straight line of charge, force varies inversely with the circumference of an imaginary cylinder (1/r).
* For any distance, r, from a plane of charge, force is constant, consistent with the idea that the flux lines are parallel.

The dilemma was that empirical equations describing the first two observations contain no dependency on PI, as should be the case for the surface area of a sphere or the circumference of cylinder. Even worse, the equation describing the third observation contains the factor, 4PI, even though the Gaussian surface is flat.

The solution, of course, was to factor the reciprocal of the electrostatic constant into 4PI times an imaginary constant, epsilon_zero.

$\epsilon _{0}=\frac{1}{4\pi k_{e}}=\frac{e^{2}}{4\pi r_{e}m_{e}c^{2}}$

#### Permeability of free space--the speed of light from sleight of hand

To whom the relationships between permittivity, permeability, and the speed of light have always seemed, as Piet Hein put it in one of his famous Grooks, "A bit beyond perceptions reach ...": this section is for you.

But first, a quick review of previous results.

The electrostatic constant has been defined as:

$k_{e}=\frac{r_{e}m_{e}}{e^{2}}c^{2}=\frac{1}{4\pi \epsilon _{0}}newton\cdot meter^{2}/coulomb^{2}$

The force per unit length between parallel electron currents is:

$\frac{dF}{dL}=\pm 2 \frac{r_{e}}{d}m_{e}\dot{Y}\dot{Z}$

With current expressed in amperes, and d expressed in meters, the above equation becomes:

$\frac{dF}{dL}=\pm 2\frac{k_{e}}{c^{2}}\frac{I_{Y}I_{Z}}{d}=\pm \frac{2}{4\pi \epsilon _{0}c^{2}}\frac{I_{Y}I_{Z}}{d}$

Defining a new constant, μ0, such that:

$c^{2}=\frac{1}{\epsilon _{0}\mu _{0}}$

The force per meter between parallel (ampere) currents becomes:

$\frac{dF}{dL}=\pm \frac{\mu _{0}}{2\pi }\frac{I_{Y}I_{Z}}{d}$

Since,

$\epsilon _{0}=\frac{1}{4\pi k_{e}}=\frac{e^{2}}{4\pi r_{e}m_{e}c^{2}}$
and,
$\mu _{0}=\frac{1}{\epsilon _{0}c^{2}}$

$\mu _{0}=\frac{4\pi r_{e}m_{e}}{e^{2}}=4\pi \times 10^{-7}newton\cdot second^{2}/coulomb^{2}\sim newton/ampere^{2}$

In the past, μ0 has commonly been referred to as the permeability of free space, but recently (2019) Standards Organizations have announced that magnetic constant is the preferred name for μ0, recognizing that it is not a property of space as the old name inferred.

The official definition of μ0 has also been changed to:

$\mu _{0}=\frac{2\alpha }{e^{2}}\frac{h}{c}$

(Alpha is the fine structure constant, and h is Planck's constant)

#### 861 : The forbidden number

The new official definition of μ0 brings into sharp focus a bias within the accredited theoretical physics establishment. It is a bias evident from the exclusion of any reference to either the classical electron radius or the Compton wavelength from descriptions of fundamental interactions. This, despite the fact that these two lengths define both electric potential energy and photon energy.

In the first case, the absolute potential energy between unique pairs of quantum electric particles is:

$\left | E_{p} \right |=\frac{r_{e}}{r}m_{e}c^{2}$
In the second, the energy of a photon is:

$E_{\phi } =\frac{\lambda _{C}}{\lambda }m_{e}c^{2}$

Beyond the deprecation of these two fundamental lengths, their relationship to each other has been obscured:

$\lambda _{C}=861r_{e}$

The above equation is not recognizable to most physicists until both sides are divided by $\inline 2\pi$:

$\frac{\lambda _{C}}{2\pi }=\frac{861}{2\pi }r_{e}=\frac{r_{e}}{\alpha }$

The left side of the equation is the reduced Compton wavelength, while the reciprocal of 861/$\inline 2\pi$ is the fine structure constant, alpha.

Planck's constant can be written as:

$h=\lambda _{C}m_{e}c=\frac{2\pi }{\alpha }r_{e}m_{e}c$
So that:
$2\pi r_{e}m_{e}=\alpha \frac{h}{c}$

The previously derived expression for μ0,

$\mu _{0}=\frac{4\pi r_{e}m_{e}}{e^{2}}=4\pi \times 10^{-7}newton\cdot second^{2}/coulomb^{2}\sim newton/ampere^{2}$

, can easily be converted to the new preferred definition:

$\mu _{0}=\frac{2\alpha }{e^{2}}\frac{h}{c}$

While this new definition of μ0 might appear computationally challenging, it can be simplified by expressing e2 as:

$e^{2}=\frac{\alpha \hbar}{c\times 10^{-7}}coulomb^{2}$

Substituting e2 into the new definition demonstrates its equivalence.

$\mu _{0}=4\pi \times 10^{-7}newton/ampere^{2}$

#### Magnetic force in the center of a circular current loop

The static repulsive force between two point concentrations of electron charge (in coulombs) can be expressed as:

$F=\frac{r_{e}m_{e}}{e^{2}}\frac{c^{2}}{r^{2}}Q_{Y}Q_{Z}$

The previous analysis of the magnetic attraction and repulsion between parallel and anti-parallel currents suggests that magnetic force might be accounted for by the following modification to the static equation:

$F=\frac{r_{e}m_{e}}{e^{2}}\frac{(c^{2}-V_{Y}\cdot V_{Z}) }{r^{2}}Q_{Y}Q_{Z}$

To test this hypothesis, consider a single conductor arranged in a circular loop. Assume a clockwise flow of equally spaced electrons, all moving with a tangential velocity, VZ. We will calculate the net magnetic force of this current loop on a concentration of QY coulombs of electrons at the center of the loop.

Before diving into the calculus, however, there are some general properties of the dot product that will simplify the integration.

First, note that there is no magnetic contribution of force if either velocity (VY or VZ) is zero.

Second, since the electron flow is clockwise, the dot product is positive for any VZ on the left side of VY, and negative for any VZ on the right side. The minus sign in the equation infers that QY will be attracted to the left side of the current loop and repelled by the right side of the current loop. (Left or right is from the perspective of QY relative its direction, VY.)

Since both semi-circles of current cause equal imbalance of the static equilibrium, we only need to integrate the magnetic force from half of the current loop, then double it.

Now consider the two quadrants of current on the left side of VY, 'left-aft' and 'left-forward'. There is a magnetic attraction to all parts of both of these quadrants. Components of magnetic attraction parallel to VY are drag forces from the 'left-aft' quadrant, but thrust forces from the 'left-forward' quadrant. These drag and thrust forces will cancel each other from symmetric positions on the two quadrants. Therefore, there is no net accelerating or decelerating magnetic force from the current loop.

This simplifies the integration in two ways:

First, we now need to integrate only one quadrant and multiply times four to find the total magnetic force from the current loop.

Second, we only need to consider the components of magnetic force perpendicular to the velocity, VY.

The quadrant of integration will be 'left-forward', with phi being the angle between the left perpendicular to VY and any differential charge, dQZ. The perpendicular component is the radial magnetic force from any dQZ multiplied times the cosine of phi. Multiplying again times the cosine of phi converts the product, VY VZ, into the dot product. $\left ( V_{Y}\cdot V_{Z}=V_{Y}V_{Z}\cos \phi\right )$

The total magnetic normal force acting on the trajectory of QY is, therefore:

$F_{\perp }=\frac{r_{e}m_{e}}{r^{2}e^{2}}Q_{Y}V_{Y}V_{Z}\frac{dQ_{Z}}{d\phi }4\int_{0}^{\pi /2}\cos ^{2}\phi d\phi$

Since,

$I_{Z}=V_{Z}\frac{dQ_{Z}}{dl}=V_{Z}\frac{dQ_{Z}}{rd\phi }$

and,

$\int_{0}^{\pi /2}\cos ^{2}\phi d\phi =\frac{\pi }{4}$ ,

$F_{\perp }=\pi \frac{r_{e}m_{e}}{re^{2}}I_{Z}Q_{Y}V_{Y}$

Force only exists between QY and QZ if both charge concentrations are constrained from accelerating towards or away from each other. In the arrangement being analyzed, only charge QZ is constrained to a circular path in the reference frame of the observer, while QY is free to accelerate.

Unconstrained mass always moves in a force-free inertial trajectory, i.e., what Einstein labeled a "geodesic path". Therefore, the trajectory of QY will curve at the angular rate, $\Omega$, required to neutralize the normal force.

$F_{\perp }=\pi \frac{r_{e}m_{e}}{re^{2}}I_{Z}Q_{Y}V_{Y}=Q_{Y}\frac{m_{e}}{e}\Omega \times V_{Y}$

$\therefore \Omega =\pi \frac{r_{e}}{r}\frac{I_{Z}}{e}$

Note that the angular rate of turn is independent of VY. Since any velocity, VY, has the same turn rate, the local inertial reference frame of QY can be considered to be rotating at the angular rate, $\Omega$.  From our non-rotating reference frame the observed angular rate of the trajectory, (e.g. through a cloud chamber), will be the Coriolis angular rate of $2\Omega$. (The factor of two comes from the fact that the normal acceleration, (Omega-cross-velocity), is relative to the local reference frame that is itself rotating at angular rate, Omega.)

While Coriolis acceleration has been deprecated as being a mere illusion, it results in a very real force if the trajectory is constrained in the reference frame of the observer.

If QY is constrained by a stationary conductor, the magnetic side force will be measured as:

$F_{\perp }=2\pi \frac{r_{e}m_{e}}{re^{2}}I_{Z}Q_{Y}V_{Y}=\frac{\mu _{0}}{2r}I_{Z}Q_{Y}V_{Y}=Q_{Y}\left ( V_{Y}\times B \right )$

$\therefore B=\frac{\mu _{0}}{2r}I_{Z}$

B is perpendicular to the plane of the current loop. By convention, the direction of B is given by the right hand rule for a positive current. For the clockwise electron current of this analysis, B points up, and the cross product points right. But since QY represents a concentration of electrons (negative charge), the force is to the left.