# The electromagnetic forces for solar energy

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### banerjee...@gmail.com

Sep 29, 2020, 2:44:59 AM9/29/20
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The Models and Calculations of Solar Energy without fusion – 2

The calculation of the solar energy from purely gravitational effects does not have anything to do with the constitution of the core – whether it is hot, or cold. Similarly, the calculation of the solar energy from purely electromagnetic effects does not, strictly speaking, depend upon the constitution of the core. We will find it from the measurement data relating to the strength of the magnetic field in the photosphere; the density of ions and their speed.

As we will see, solar energy is mainly of electromagnetic origin. The main reason for the energy of our Sun is the Sun’s magnetic field interacting with the ions (charged particles) in the photosphere. Just as the Earth appears to have a large bar magnet inside it (with the North of this magnet near the Geographic North Pole, and the South of the magnet near the Geographic South Pole), the Sun too appears to have a huge bar magnet inside it. From these poles, magnetic lines of force spread out all over the Sun, reasonably uniformly save – very importantly – for the magnetic spot regions where they are far more intense.

We will now see what is known as magnetic field strength, also known as magnetic induction. It is a vector, which means that it has a direction as well as a magnitude, and is represented by the letter B. Consider all the magnetic lines of force (usually called magnetic flux) of the Sun, cutting into the photosphere region. These are all parallel lines, going from the Magnetic North Pole of the Sun to the Magnetic South Pole of the Sun. Any cross section of the entire photosphere region, that is, a great circle with a very large hole in it, is the surface region of the Sun’s total magnetic flux through the photosphere Call it A. Let the total magnetic flux by called, Sun-Flux. Then the average B at any point in the photosphere of the Sun is Sun-Flux divided by A; its direction is the direction of the magnetic flux at any point. Its value is given as 1 gauss; which is double that of the Earth’s magnetic field strength, or magnetic induction.

Just as an uncharged particle is attracted by a gravitational field, and accelerated by gravitational force; a charged particle (ion) is affected by a magnetic field, and accelerated by electromagnetic force, when moving with respect to the magnetic lines of force in the magnetic field. However, there is a difference. The uncharged particle moves in the direction of the gravitational field – thus a ball drops straight down to the ground by gravity, does not move sideways. But for the charged particle moving in a magnetic field, the direction of its acceleration is perpendicular both to the magnetic field and the direction of its motion. In short, it moves sideways, not down as is the case for gravity.

An ion (electron or proton) that is moving in the direction out of the Sun, will be affected by the magnetic lines of force that are running parallel to the Sun. The force upon it will act sideways – making it veer away from its original outward path by accelerating it toward a sideways, parallel path to the Sun. Thus a recently ionised hydrogen atom, an electron and proton that is, will have its two components accelerating in opposite directions. All ions moving up will behave like that; now all ions moving down will have their electrons going towards protons and protons going towards electrons. They will combine, once again to form the hydrogen atom, and in this arrest of its kinetic energy, give out radiant energy from the excited newly formed hydrogen atom. Prior to this, the ion may have collided with hydrogen atoms, causing ionisation with formation of radiant energy, and also radiant energy, part of which is visible light. If the movement of any ion is parallel to that of the magnetic induction vector, then no force will act upon the ion. But at any other angle, the force will be kept. Incidentally, this is the way the ions in the ionosphere of the Earth or Sun are kept attached – the ions that are not very fast are moved aside by the magnetic field strength, so that they remain parallel to the Earth (or Sun). However, ions (protons and electrons) do escape from the Sun, especially at the times of great solar flares when they get too much outward velocity.

By now our model for the Sun’s energy is getting clear – the magnetic field of the Sun accelerates the electrons and protons (formed by the ionisation or electron deprivation of the hydrogen atom) so that they smash against atoms causing ionisation and radiant energy, before combining again to form hydrogen atoms. This is a cyclic process – ions with increased kinetic energy smash into atoms, creating more ions, before recombining into atoms. This process would not have worked had there not been a magnetic field to accelerate the ions. No loss of mass is involved. We note that the charge on the electron is in magnitude the same as the charge of the proton. These are oppositely charged. This means that under the magnetic field they go off in opposite directions – they do not cancel each other as may appear in mathematics as they have different signs. To recombine, a proton thus created from a hydrogen atom must meet an electron created by the ionisation of another hydrogen atom.

From the physics textbook, Halliday Resnick Vol 2, Chapter 33-2 “The Defintion of B” we go straight into the formula for force upon a charge in a magnetic field. I will be quoting from the book, below.

“Let us fire a positive test charge q with arbitrary velocity v through a point P. If a sideways deflecting force F acts on it, we assert that a magnetic field is present at P and we define the magnetic induction B of this field in terms of F and other measured quantities.”
“if we vary the direction of v through point P, keeping the magnitude of v unchanged, we find, in general, that although F will remain at right angles to v its magnitude F will change. For a particular orientation of v, and also for its opposite orientation –v, the force F becomes zero. We define this direction as the direction of B.”
“The magnitude of the magnetic deflecting force F… is given by
F= q * v * B * sin(theta),
where theta is the angle between v and B.”

Let us see how the above formula can help us to show whether or not the electromagnetic force is responsible for most solar energies.

The force acting upon a proton will be thus
F= q(proton) * v * B * sin(theta).
There will be very many protons as we shall soon see, and let us assume that at the time of their ionisation there was this unknown velocity v which let us hold is the average velocity for all protons. Also there is a sin(theta) value which should be 45 degrees for the average, as half the protons will be clustered statistically around the direction of B, and the other half perpendicular to it.
There will also be the force acting upon the electron, which is
F = q(electron) * v * B * sin(theta).
Force by Newton’s second law of motion is mass * acceleration. So the acceleration of the proton due to the electromagnetic force upon it is
Acceleration of proton (Ap) = q(proton) * v * B * sin(theta)/(mass of proton), and similarly
Acceleration of electron (Ae) = q(electron) * v * B * sin(theta)/(mass of electron).
We find from above that the acceleration of the electron is many orders of magnitude higher than the acceleration of the proton, as the electron is correspondingly lighter. The electron will move much faster. If it cannot combine with a proton in the photosphere, it will escape into the Sun’s ionosphere, where, if it will meet a proton it will recombine to form a hydrogen atom and descend to the photosphere. Or else, be part of the ionic drift. The high velocity of the electron thus does not contribute to solar power as such – that is, not the way that proton-hydrogen collision does for the energy we get.
In time t, the proton under this force alone, with no consideration for any initial velocity (for conservative reasons, as we are only considering electromagnetic force impact) moves a distance Sp according to the formula
Sp = 0.5 * Ap * t^2.
In one second, the proton will move a distance Sp = 0.5 * Ap. Similarly in one second the electron will move a distance Se = 0.5 * Ae.
The energy from electromagnetics obtained by the proton in one second is the force multiplied by the distance over which the force acts. It is
Energy of proton in one second, or power from proton = (mass of proton)*Ap*Sp.
Similarly for the electron
Energy of electron in one second, or power from electron = (mass of electron)*Ae*Se. This power of the electron is very much more than the power of the proton! Our task here is to show that fusion at the core of the Sun is not responsible for the Sun’s energy. To prove our task, let us first ignore the power from the electron as the cause of the solar energy. We will, however need it to explain solar flares, novae and supernovae.
Power from the ionised hydrogen atom, neglecting the impact of the negatively charged electrons is Force multiplied by distance per second, or mass of proton multiplied by its acceleration and the distance covered in one second by that force. Call it Piah, so
Piah = (mass of proton)*Ap*Sp.
Substituting the values of Ap, Sp and simplifying
Piah = (mass of proton) * (q * v * B * sin(theta)/(mass of proton)) * 0.5 * (q * v * B * sin(theta)/(mass of proton)). or,
Piah = 0.5(q*v*B*sin(theta))^2/(mass of proton)
Now let us put in some data values here, obtained from books and internet. Mass of proton and its charge are well known. They are 1.67 * 10^-27 Kg and 1.6 * 10^-19 Coulombs respectively. With theta held at 45 deg, sin(theta) is 0.707
The value of B is 1 gauss, and in MKSA it is 10^-4 Tesla. The link for the quote is given below.
https://www.windows2universe.org/sun/sun_magnetic_field.html#:~:text=The%20Sun%20has%20a%20very,Earth%20(around%200.5%20Gauss).
The Sun has a very large and very complex magnetic field. The magnetic field at an average place on the Sun is around 1 Gauss, about twice as strong as the average field on the surface of Earth (around 0.5 Gauss).

Piah (watts) = 0.5 * (1.6 * 10^-19 * v * 10^-4 * 0.707))^2/1.67 * 10^-27, or
Piah (watts) = 0.5 * v^2 * (1.132 * 10^-23)^2/1.67 * 10^-27 = 0.5 * v^2 * 1.28 * 10^-46/1.67 * 10^-27, or
Piah (watts) = 3.8 * 10^-20 * v^2

We now find out how many protons are there in the photosphere.
The mass of the photosphere has been found in the preceding section to be (assuming a depth of 400 Km and a density of 3 * 10^-4 Kg per cubic meter to be 3.65 * 10^20 Kgs.
Let i be the fraction of the hydrogen in the photosphere that is ionised.
Then the total mass of the ionised mass in the photosphere is i * 3.65 * 10^20 Kgs.
Going by the fundamentals of Physical Chemistry, in 1 gram of Hydrogen – a mole - there are N molecules of Hydrogen, where N is the Avogadro’s Number, 6.02 * 10^23. In one Kg of Hydrogen there are then 1000 * N atoms, or 1000*N protons when their atoms are ionised (deprived of electrons).
The total number of protons in the photosphere (Npp) is thus 1000*N*(i * 3.65 * 10^20) units. Or i * 3.65 * N * 10^23, or i * 21.93 * 10^46, rounded to
Npp = i * 2.2*10^47

Now the unknowns are i and v. Let us see what a ball-park value of i (the degree of ionisation in the solar photosphere could be). Internet search gives us this link:
https://astronomy.stackexchange.com/questions/7883/what-is-the-degree-of-ionization-is-the-solar-photosphere
where it is written:
“The book "Introduction to Stellar Astrophysics" by Boehm-Vitense, vol. 2, 1997 reprint, on page 76 claims that one out of every 10^4 hydrogen atoms are ionized, so hydrogen is mostly neutral in the photosphere;”
In the same link we have a more detailed reference
“The degree of ionization in the photosphere varies with depth of course, but overall it is small. Table 1 of the Bilderberg Continuum Atmosphere (Solar Physics, 3, 5, 1968) gives the pressure and the electron pressure at various optical depths in a comprehensive model. The ratio of the pressure gives the Ne/Ntotal. At optical depth = 1 at 5000 Angstroms, Pgas = 1.412E5 (cgs) and Pe = 6.239E1. The ratio is 4.4E-4. The hydrogen ionization, also given in the table is 4.07E-4.”

We will take, for the cause of conservativism, the lower value of 10^-4 for the value of hydrogen ionisation in the photosphere. Then the number of protons in the photosphere will be:
Npp = 2.2 * 10^43.
The total solar energy from the photosphere, caused by proton movement alone, will then be Npp * Piah = 2.2 * 10^43 * 3.8 * 10^-20 *V^2 = 8.36 * 10^23 * v^2 watts.
Now, what is left is to find the value of v, which is the velocity of the proton in the direction perpendicular to the magnetic lines of force. Its absolute velocity is not the value required. Internet searching in “Google Scholar” for the value of the velocities of the atoms in the photosphere shows us in the link https://link.springer.com/article/10.1007/BF00153898
for the paper “A physical mechanism for the production of solar flares” gives a figure of greater than1 km/second for the velocity of the gases in the photosphere. Other sources give much higher values. These values are absolute values – their component that is perpendicular to the magnetic lines of force is not known.

Taking the value of v as 1Km/second or 1000m/s and putting it in the above equation we get the value of total solar energy from the photosphere to be 8.36 * 10^29 watts.

Hey! The actual total solar energy as calculated (and verified by experiment) is only 3.864 * 10^26 watts. And of that only 3.5% is from gravitational forces. Our calculation above gives the electromagnetic sources for solar energy over 2000 times the actual solar energy. We have been very conservative – we have not taken the energy from electrons for the energy from the photosphere; we have taken a low value for the ionisation of the photosphere. Some sources say that the photosphere is only 100Km in depth, not 400Km as we have taken. Even when we take the former value of 100Km, the energy from electromagnetic forces would still be 500 times the actual energy.

This does amount to an overkill for my proposal that the solar energy is NOT from fusion at the core. Of that, there can be no doubt. So my main thesis, that the Sun has a cold iron core, where circulate huge currents from superconductivity, causing the magnetic field, which moves the ions to create the bulk of the solar energy, is validated.

The huge value of 8.36 * 10^29 arises from our taking the value of v to be 1000m/s. This v can never be known by measurement, for we cannot experiment on the Sun, but it can be calculated from measurement. What is it? Well, we have found in the earlier section that the solar energy from gravity alone is 13.7 yotta watts. The solar energy not from gravitational forces is then (386.4 – 13.7) yotta watts or 372.7 yotta watts. This then should be equal to the power from the electromagnetic forces. Taking the photosphere to be 400 Km deep, and with i=10^-4, B=10^-4 teslas we get the relation
8.36 * 10^23 *v^2 = 372.7 * 10^24.
From where v^2 = 3727/8.36 or v = 21 m/s

If the photosphere were 100 Km deep, there would be one fourth the amount of protons, in which case the value of v would be only double at the very reasonable value of 42 m/s. Thus it seems fairly certain that solar energy need have nothing whatsoever to do with very hot fusion processes in the core of the Sun. That the core of the Sun, like any other star, is very cold, and unaffected by whatever going on around it, is crucial for our understanding of novae and supernovas.
In the next section we will consider the impact of the fast moving and powerful electrons for creating solar flares.

Arindam Banerjee
Melbourne 26/09/2020