MAXWELL EQUATIONS WITH E.M

(This article comprises of 3 parts: The first is, almost about the curl operator (likely a copy from manual book) and a half of Maxwell differential equation; the second and third parts are totally written by author. An important note is that the equation can be solved only with time and place; furthermore almost every quantity in the problem is vector. You can understand and debate more after reading through.)

I-                 Maxwell differential equation:

The equation set is named after the physicist and mathematician James Clerk Maxwell.

Curl of B or

is a differential equation in the set of following several equations:

Figure 01-Set of Maxwell equations

From Maxwell-Faraday to Maxwell’s addition on Ampere, the relation between magnetic field and electric field is formulated. The consequence is that a magnetic field induces the electric field to change or vice versa. This is fundamental to electromagnetic wave theory, and by such pattern the electromagnetic wave is transmitted.

By mathematics manual, the equation is solved as following:

If a given ‘’B’’ is

The vector quantity ‘’B’’ is almost unique to any position in the world.

The curl of B in the Cartesian coordinate is:

Where unit vector ‘’x’’ indicates that the value before it is on ‘’x’’ wise, or( is on ‘’x’’ direction or illustrating the functioning of By and Bz on ‘’x’’. It is ditto to ‘’y’’ and ‘’z’’ for each of 2 other directions.

The ‘’Curl of B’’ may be understood as B gradient with compositions on ‘’x’’, on ‘’y’’ and on ‘’z’’.

Since the mid-20th century, the Maxwell's equations are realized as giving out un-accurate result, ‘’but a classical limit of the fundamental theory of quantum electrodynamics.’’ An example of non-accurate result is its application to Earth Magnetism.

Notwithstanding, for the sake of explaining 24-hr E.M revolution, Ampere’s circuital law (with Maxwell’s addition) in differential equation is applied. That equation is to be reviewed.

 II-Problem of 𝜵*B=µ₀*(J +Ɛ₀* δE/δt):

1-Twin graph of J and gradient of E:

The following graph is built out of real data that observed at Tirunelveli (South India). Both plots on this graph (gradient of atmospheric electric potential ‘’E’’ and density of air-earth current‘’J’’) are averaged in every minute; one is in proportion with the other (fig. 02).

Figure 2- The air-earth current ‘’J’’ is averaged in every minute, observed with Wilson antenna(broken line).The continuous line is performing the gradient of atmospheric potential that measured with passive antenna at Tirunelveli on 25 April 2007 (reference to Atmosphere Electric of NOAA).

After every 24hr interval, both ‘’J’’ and ‘’E’’ (refer to the graph of figure 2) vary and especially their fluctuation repeat (24hr) on global scale except at locales where some specific matter effects on those 2 figures.

We may ask whether the fluctuations of ‘’J’’ and ’’E’’ contribute to Earth Magnetism and explained by Ampere-Maxwell equation?

Definitely the answer must be ‘’yes’’, and it is right in the Wikipedia for ‘’Ampère's law with Maxwell's addition the Ampere rule with Maxwell addendum says that magnetism can be created by both electric current(Ampere) and changing electric fields. This statement affirms that the fluctuations of both ‘’E’’ (potential gradient) and ‘’J’’ displacement current are inducing Earth Magnetism.

For Earth Magnetism, the contribution from atmospheric electric is obviously as mentioned above. But we need more argument anyway, if the terrestrial magnetic field is belonging to one of an individual magnetic bar, the above mentioned reason for Earth Magnetism ‘’Electric field change’’ may be accepted as the major; or implies: the atmospheric electric potential and displacement current vary to induce Earth Magnetism.

Actually the Earth Magnetism is a set of fields that induced by the moving of many electric charges, definitely neither just the fluctuations ‘’J’’ and ‘’E’’ nor a field of such a powerful magnetic bar inside the Earth core. But the Earth is magnetized by magnetic fields and then the fields expand outward from the Earth.

Notwithstanding we do carry out an express review to Ampere-Maxwell once as to consider ‘’J’’ and ‘’E’’ in the graph on figure 2.

General description in brief:

-The homographic plots are built out of real data that observed at Tirunelveli (South India). Both two quantities, air-earth current and gradient of atmospheric voltage, are fluctuating, and its fluctuation is repeated after every 24 hrs.

-The horizontal line is for time that calibrated in 60 minutes (1hr.)or 1440 minutes in 24 hrs.

-Assumption: Each 60 minute segment equals to 1 segment of 10 V/m on vertical line that performs gradient of atmospheric voltage, meantime equals to ½ of a segment on another vertical line which performs the density of air-earth current.

The world scientists spend a lot of time to debate about this problem, eventually they all recognise what happened as realities. I do explain about this question in ‘’Chapter I’’ of the book: ‘’the Earth Magnetism and some basic problems with sea-land surface, atmosphere, Van-Allan bell and Moon’’. The explanation in there also implies that the above graph may changes a bit in accordance with the change of position of the Sun relative to the Earth.

-The above graph is to apply worldwide because the data are recognised as the same for everywhere in the world. 

-There are always, within 24 hrs., 2 moments when air-earth current almost reaches zero while potential gradients are still hung at 100V/m or a bit better (the minute 175^th and minute 785^th). The plots are built of real data, therefore they don’t look similar to any standard graph such as Sin, Cos or Tg...

-There are many parts of those plots likely straight, some of the straight line is forward slash while some other is backward slash.

Figure 04- ‘’J’’ and ‘’E’’plots-Red colour column indicate‘’rising up’’,

while the grey is for falling down.

My way is to attribute a certain straight line to each part of ‘’E’’ plot. Seek for specific differentiation of potential gradient of ‘’E’’ on ‘’t’’ (time) in that part (  ).

Some tiny problem will be setup and sorted out during Q&A.

Q&A:

Q1-When does 𝛛E/𝛛t reach 0 in each 24hr interval?

Throughout a 24hr interval, the ‘’E’’ and ‘’J’’ graphs are rising up and falling down many times, the values of 𝛛E/𝛛t at every peak and every bottom are zero because where the tangent line are horizontal or parallel to (t) line.

Figure 05 –At every peak or bottom, the value of 𝛛E/𝛛t may reach Zero

Approximately 𝛛E/𝛛t may reach zero value at the following times 60, 90, 180, 400, 510, 640, 785, 1085, 1145, 1265^th minutes.

When 𝛛E/𝛛t reaches zero, the value of 𝜵*B may be considered as function of ‘’J’’ only:

𝜵*B=µ₀*J

As matter of facts, in the circumstances when 𝛛E/𝛛t=0, when 𝜵*B= µ₀*J; the value of ‘’J’’ plays no important role, while the trend of electric field does. Refer to figure 04 above, right after minutes 179-180th, or at 510^th, 785^th, 1145^th the value of 𝜵*B or gradient of B almost rocketing of falling down sharply.

Q2-The value of 𝜵*B at minute 960^th GMT (The popular slope within 24hrs.)?

On figure 4, the minute 960^th is belonging to a red part of the graph, it is rising. To solve this problem, we need the homographic graph for ‘’J’’ and ‘’E’’ (From NOAA), with some additional lines on it.

+We can find the value of ‘’J’’ at minute 960^th by drawing a vertical-straight line through 960, we get Nand M when it meets the plot of ‘’J’’ and ‘’E’’.

From N, draw a horizontal line parallel with time line, it meets O’E’ at K; Position of K on O’E’ is to indicate value of ‘’J’’ which is approximately 0.62*10^-12 Amp/m2.

+From M, graph of gradient E is approximately on a line that meets time line at D(approximately it is t=730). The line in 2D coordinates should be:

(y=ax+b).

+In order to find out 𝛛E/𝛛t, we consider a function of ‘’E’’ in the form y=ax+b at minute 960 GMT.

E=at+b (‘’t’’ is short for ‘’time’’), a=tg (MDH)) (This function is set up for M and its adjacency only).

𝛛E/𝛛t = a (Applied for the function of E=at+b).

Figure 6-Gradient E and displacement J with some additional lines at minute 960^th.

The point M is chosen at a positive part of (forward and popular slope) of 24hr plot of gradient ‘’E’’.The line MD coincide with a part of the above plot and its function is: y=ax+b.

We got to set and solve a problem.With t=960, E=Em≈168 V/m and whent=730^th, E=80 V/m. The results are:

a≈(168-80)/(960-730)=0.3826 and b≈-199.296

Thus, the function for E at t=960^thcan be:

E=0.3826t - 199.296

Value of 𝛛E/𝛛t=0.3826 (indicates the slope of the line tangent to a curve at the intended point). The function for E as mentioned is applicable to the point M and its adjacencies only. For other parts of that plot, both ‘’a’’ and ‘’b’’ will be attributed with some different values.

With ‘’J’’=0.62 pA*m^-2 at t=960 (1 pAmp/m2=10^-12 Amp/m2, µ₀=4π*10^-7, Ɛ₀ =5*10^-15 (average between 3*10^-15and8*10^-15. Furthermore, we should note that 1 pA=10^-12 Ampere).

𝜵*B=µ₀*(J +Ɛ₀* )=4π*10^-7*(0.62*10^-12+5*10^-15*0.3826)=

=12.56637*10^-7*(620*10^-15+5*10^-15*0.3826)=12.56637*10^-7*621.913*10^-15

=7.81519*10³*10^-15*10^-7=7.81519*10^-19

𝜵*B=7.81519*10^-19

This is value of 𝜵*B at minute 960^th GMT on daily basis.

-The value of 𝜵*Bat minute 960^th GMT is found as an example that indicates that we are capable to find out value of 𝜵*B at any time in a day. As long as ‘’E’’ and ‘’J’’ fluctuate with 24hr revolution, the value of 𝜵*B does the same, and it can be considered as daily revolution of Earth Magnetism.

Q3-When and where we find 𝜵*B<0?

𝜵*B=µ₀*(J +Ɛ₀* δE/δt)

Let’s solve this problem:

If (J +Ɛ₀* )<0

Then < -J/Ɛ₀

When the value of ‘’E’’ is falling down, definitely  δE/δt  <0, but this problem requires a certain ‘’J’’.

The quantity ‘’J’’ varies within 0 and 1.25*10^-12, while Ɛ₀ does not change often, we supposed that

Ɛ₀ is average between 3*10^-15 and 8*10^-15, and assumed as 5*10^-15.

When ‘’J’’ is rather high, let’s consider J=1.25*10^-12 Amp/m² in our example, then: Then ≈-250, this is a slope of the ray that nearly vertical downward (89⁰ 43’).

If we consider ‘’J’’ at sometimes when its value is low, such as the duration from minute 150^th to minute 160^th, J ≈  0.05*10^-12 Amp/m2. The same calculation as above is applied, we obtain:

≈-10, equivalent to tg of an angle 84⁰15’ downward.

Figure 6a-Angle of 84⁰15’ downward.

In brief, we can realize that the lower ‘’J’’ the more possible for 𝜵*B<0. Nonetheless, no matter how much the value of 𝜵*B gets lower under zero, this is to indicate that the amplitude of Earth Magnetism is getting down due to the fluctuation of displacement current ‘’J’’ and potential gradient ‘’E’’, they expect to be the same in the next day; and nothing else about the average of Earth Magnetic Magnitude.

Anyhow, this is not more than a calculation for 𝜵*B,some more attract our attention are the sources that induce the Earth Magnetism, someone are from the electric charges on the Earth and in the atmosphere;while some other ones are in Near Earth Space.

Q3-How is the influence from potential gradient ‘’E’’ and displacement current ‘’J’’ to Earth Magnetism?

By Maxwell, as matter of facts, the variation of electric field can induce the magnetism. We even do some review for questions 1 and 2. However, for a better pragmatic concept, another figure is to be considered: The fluctuation of total area of thunderstorm in the world.

Figure 7–The fluctuation of global thunderstorm area (km2).

Almost every electric and magnetic incident in the world is created and developed by‘’J’’ and ‘’E’’ as discussed in several paragraphs before, and even in this essay. But the conspicuous subject is the total area of thunderstorm in the world, the figure 7 above is an important reference from NOAA.

In general, the fluctuation of total thunderstorm area in above graph is explained as consequence of ‘’E’’ and ‘’J’’ fluctuation, the displacement current ‘’J’’ variation makes the value of charge on Earth surface changed, and then one place is different from the other. Consequently, the charge on land surface contributes to Earth Magnetism (Chapter I-Air-Earth current and Chapter II-Contributions from Land & Sea surfaces to Earth Magnetism by Biot-Sawart laws.)

The move that makes the electric charge on the Earth surface changed in order to change the Earth Magnetism at all time.

(Problem (this can be a reference to student or lecturer): The contributions from sea surface and land surface to Earth Magnetism at 13hrs GMT? If The Earth is neutral or total charges on land surface and sea surface is null, every land square km under thunderstorm cloud is charged with 10⁶ Coulomb.)

Q4 – Is any electromagnetic wave created from atmosphere electric?

Electric field fluctuation is to induce magnetic field and vis-a-versa. This is traditional electromagnetic theory in science, it is also fundamental theory of electromagnetic wave transmission, and it is known to the world.

By the theory mentioned above, the fluctuation of electric field described in figure 2 can be viewed as an oscillation which produces one or several electromagnetic waves. The waves are emitted into universe with such a huge power that nothing can stop or consume them all…

The electromagnetic wave detectors on board spaceship can realize these waves from Earth.

Figure 8-Basic frequency (Left)and the form of original wave (Right) incorporated in 1 coordinate for comparison(Cited from Fourier manual)

All waveforms, no matter what you scribble or observe in the universe, are actually just the sum of simple sinusoids of different frequencies.(Fourier)

My own speculation, the 2 obvious waves can be detected from outer space are wave of T≈k1*640’, and T≈k2*1600’are popular with huge output regardless the capacitor-coil are found or not.

Figure 9 – Principal description of electromagnetic wave creator.

CONCLUSION:

-Although the homograph of E and J are applied everywhere in the world.

-Although the differential equation of Ampere-Maxwell does not bring us an accurate outcome for Earth Magnetic magnitude at any time. But the gradient of Earth Magnetism (𝜵*B=µ₀*(J +Ɛ₀*δE/δt)) at any time is found throughout every 24hrs to affirm that: if revolution of the two quantities (E and J) is 24 hr. time, then the revolution of Earth Magnetism revolution is 24hr too.

There are several magnetic inducers to Earth Magnetism found, therefore every calculation depending on functions of atmospheric electric field and displacement current can’t bring satisfactory result. So, we can’t claim Ampere-Maxwell equation a perfect instrument for predicting the current value of Earth Magnetism.

Another reason for above negative comment is that the Earth Magnetic vector does not just vary in time, but also at location where the vector is observed; while Ampere-Maxwell may give out a result that depends only on GMT.

-Although the fluctuation of ‘’E’’ and ‘’J’’ –as above considered with Ampere-Maxwell equations--may influence to Earth Magnetism ambiguously. But consequence of the fluctuation--the thunderstorm area--is an important figure that contributes enormously to Earth Magnetism and makes it varying with revolution of 24hr (discussed in chapter III).

-The Earth in its atmosphere with Ion layers establish a unit in the universe, that unit is creating electromagnetic waves and emitting them into the universe.

I hope that one day, a certain satellite in outer space will detect those emitting waves; these are created by nature and from Earth. Unlike the known waves from Earth, these mentioned waves are produced in Earth atmosphere with unknown output. The waves are produced continuously, unlimited power and non-stoppable. They are voices broadcasting from our Earth.

  

Hanoi Nov. 30th 2018
Nguyen Van Cuong
Advisor/Editor: Phd. Vu Bang

CHAPTER II/p1-SEA VS LAND

PROBLEMS OF SEA vs LAND CHARGEs

The last chapter presented a discussion about positively charged land surface and the negatively charged sea surface. They both are parts that constituted the Earth and moving together with Earth’s rotation; if the charge of each is opposite to the other, then each induces magnetism that contrarian to the other. In the paragraph 2 (Argument of electric paradox) we even discussed and mentioned the average ratio +ions/-ions in every CC of air on land.

However, we leave that ion ratio for another discussion and work with graph of thunderstorm coverage for our problem in this chapter. So we set and solve the following problems with an assumption that: total contribution of every charged cloud on the Earth is ‘’nil’’ throughout our problem solution.

I-PROBLEM OF A SINGLE CHARGE ON THE EARTH SURFACE:

1-Problem:

-A charge of -100 Coulombs, locating at a fixed position P (30N, 100E) on surface a globe model of 100 m radius. Another charge +100 Coulombs, locating at the same fixed P on altitude h=10 m above surface of globe. How is magnetism induced by those 2 charges to rotation axis if its rotation is the same as Earth’s cycle (24 hrs)?

a-Analyse:

This figure is displaying a global coordinate, where ‘’O’’ is centre and P is a given point on global surface.

The latitude of P is also the angle ‘’ϕ’’ between OP and equator’s plane.

Figure 1/II-Charge at P versus P1 on global coordinates (not in scale).

The imaginary plane whereby the P moving on is the plane of latitude slice at 30N- depicted in the next figure.

The distance from P to global rotation axis is the value of ‘’r’’ in our calculation:

PP2=r=R*cos(ϕ)

The distance from P1 to global rotation axis is another value of ‘’r₁’’ in our calculation:

P1P1’=r₁=(R+10)*cos(ϕ).

The global rotation is to carry P and P1 together. In scope of electromagnetism, longitude of P (100 E) plays no role or helpless in our calculation.

The globe is rotating around its own axis at all the time. The ‘’P’’ is fixed on globe’s surface and moving along with global rotation. Every single charge moving around a certain point is to create a loop of electric current by which the magnetism is induced. The following is Biot-Savart expression (9.1.20) to facilitate our calculation:

B=(μ/4π) * (qv*''r''/r(2))

(v is speed vector of a charge and ''r'' is unit vector of radius, the vector is not presented properly in this blog)

========Note

Permeability (μ) of the medium material between both P,P1 and globe rotation axis: As we are working with Earth model, therefore we suppose its permeability to be one of Earth, the average of Iron (99.95%) (Because Earth core is assumed as Iron), and normal air (µ0 =4π*10-7 N/A2). In reality µ is varying, the permeability at different place is different. This is only to set a way to solve the real problem. ==========

b-Assumptions: In order to solve this problem, we need some assumptions.

-The globe’s average radius R=100 m so:

 r=R*cos(ϕ)=100*cos(30⁰)=86.6 m

r₁=110*cos(30⁰)=95.26 m

-Permeability of the medium material between P,P1 and globe rotation axis: As we are working with Earth model, therefore we suppose the permeability as one of Earth, the average of Iron (99.95%) (Because Earth core is assumed as Iron), and normal air (µ₀ =4π*10^-7 N/A²). So we have: 

µ=1/2 (0.25+4π*10^-7) =0.12500126 N/A²

General formula for speed (v) of P: v=2πr/24*60*60 m/s, replace ‘’v’’ to Biot-Savart expression:

dB= *q/(r*48*60*60)=

dB=7.233869*10^-7*q/r (named as I-Biot-Savart).

Although the coefficient 7.233869*10^-7 is derived from µ, π and cycle (24 hrs), it is interpolated from Biot-Savart expression, so we name it as ‘’ interpolated Biot-Savart’’ or I-Biot-Savart for further reference.  

With I-Biot-Savart, the only dependent to location of the charge ‘’q’’ on the Earth is its distance to Earth’s rotation axis ‘’r’’, while ‘’q’’ is to be found out.

Thus, we can be proud to say: if I have a charge and its location on the Earth, we find out how much it contributes to Earth Magnetism.

c-Solution with Biot-Savart:

Solution 1:

-Given charge at P: q=-100 Coulombs

-Radius of the loop: r= 86.6 m   

-Product of vector speed ‘’v’’ and unit vector ‘’r’’ is a vector that parallel with Earth’s rotation axis, pointing to true North and valued |v|, because vector  is perpendicular to unit vector .

-Value of Δb1 is calculated step by step with the above Biot-Savart interpolated as: =7.233869*10^-7*q/r=-7.233869*10^-7*100/86.6=-8.353197*10^-7

Solution 2:

-Given charge at P1: q1=+100 Coulombs

-Loop radius: 95.26m

-Value of Δb2 is also calculated step by step with the above Biot-Savart interpolated as:

=7.233869*10^-7*q/r1=7.233869*10^-7*100/95.26=

=7.593816*10^-7

d-Summary words: (apply I-Biot-Savart).

-8.353197*10^-7 + 7.593816*10^-7=

=-0.7593816*10^-7 Tesla

If two oppositely charged points of the same true value are placed on the same position at 2 different altitudes; resulting the distance from each to the rotation axis to be different to the other and the intensity of magnetism contributed by each charge is different to the other. The closer to axis, the more it contributes; that’s rule. The total of the two is in favour of the inner charge.

II-PROBLEM OF OCEANS VS CONTINENTS

1-Basic argument:

The Sun is rather far from Earth, so it can’t apply any magnetic influence on the Earth directly but indirectly through brokerage of atmosphere, continent and ocean surfaces. Beyond the influence that almost considered in the last chapters, this is a problem for the contributions of continent and ocean (or sea & land as I name them in this book) to the Earth Magnetism in the light of Biot-Savart laws.

2-Thunderstorm matter:

From atmospheric research, the thunderstorm area varies against the time with 24-hr revolution or daily cycle. The following is a graph that re-built by the data taken at Mauna Loa Observatory (Hawaii Islands):

Figure 3/II-This is observation result from Mauna Loa.

The graph is indicating that the Americas continent is the place of thunderstorm and lightning, while the less thunderstorm continent is Asia. The average thunderstorm continent is Africa which is at the same peak as Americas.  

The above figure is divided approximately into 3 trapezia and 1 rectangle to calculate the total area of thunderstorm and its average within 24 hrs. The result is the average area under daily thunderstorm ‘’80*10⁴ km²‘’, used as entry for our problem.

Figure 4/II-Global average area under thunderstorm daily.

Thunderstorm and lightning on 30 March 2017 (as instant only, the situation is changing at all the time).

Picture 5/II-An instant image (courtesy from Blitzortung.org)

The world thunderstorm map is indicating that the thunderstorm is covering almost within 55N to 55S, where some spotted lands should be mentioned in the South semi-sphere such as Australia, Indonesia. So, we consider a reasonable way to distribute the area of 80*10⁴ km² under daily thunderstorm, make those figures the entries to our problem.

Note: (Under the thunderstorm, the area is almost positively charged and assumed to be so).

Figure 6/II-Illustrative structure of a thunderstorm cell

(The above figure is re-built after many real flights through a super cell, the pilots who performed the flight surveys were recognised as braves in atmosphere research.

The obvious sign illustrated above is the positive land surface that expands as large as the thunderstorm cell is. The positive patch on land, as matter of fact, must be laid there until all the charges above to be discharged. Furthermore, the land surface is moving during Earth rotating with as high speed as V*cos(ϕ) (V-speed of any position at equator-463 m/s, (ϕ) is position’s latitude).

3-Problem:

Daily global average area under thunderstorm is 80*10⁴ km². If the land under thunderstorm is positively charged (supposed average +1000 Coulombs/km²); reciprocally the oceans are negatively charged so that they can balance with the land positive charge (P65-Bal.).

Assume that the land average height is 10 m above mean sea level, while sea level is assumed as at its average level during our calculation.

What is the contribution of sea and land to Earth’s Magnetism?

X

X                             X

There are obviously several ways to solve the problem. Our way is to attribute a certain charge to each potential location, calculate contribution of each and tantalize at the end of this section.

The seas (and oceans) are considered as one single electrode because salt water in ocean is homogeneous with high electric conductivity. The assumed value of electric charge is distributed evenly to everywhere of seas and oceans. We will calculate the ocean contribution to earth’s magnetism and summed up with contribution from lands.

3a-Solution-Land contribution (Δ1):

Expression I-Biot-Savart (p84) for Δ(i):  

Δ(i)=7.233869*10^-7*q(i)/r(i)

Back to our problem, it should be simplified by distributing approximately the total given land charge to 6 different places; each of them represents a specific locale where the thunderstorm land is dominating its region.

The land charge is estimated on basis of daily air-earth record (figure 4/II), and every charge is positive, furthermore as noted before that each diminishes the E.M.

From entry, we have 80*10⁴ km² area and assumed 10³ Coulombs/km²; so total charge on land is assumed as 8*10⁸ Coulombs. From every position, we always find the distance to Earth rotation axis (assume that average height of land charge is 10 m above average sea water level).

With approximately estimated percentage of areas under thunderstorm, we also attribute average area under thunderstorm for each different location. Eventually an assumed charge at each location can be found. A table of mock calculation is established as following:

-Assumption1: by percentage, the areas under thunderstorm with respective charges are approximately divided as following:

+America continental (AC): 20 pcts, (35N, 90W.)

With r(1)+10=5,207,195.7m, and q(1)=16*10⁷

+North Africa & Europe (A&E): 20 pcts, (40N, 20E.)

With r(2)+10=4,869,601.3m, and q(2)=16*10⁷

+South Africa (S.A): 10 pcts, (10S, 30E.)

With r(3)+10=6,260,235.9m, and q(3)= 08*10⁷

+Japan and its’ neighbour (Jap.): 15 pcts., (35N, 140E.)

With r(4)+10=5,207,195.7m, and q(4)= 12*10⁷

+Philippine and Indonesia (P&I): 15 pcts., (0.0, 115 E.)

With r(5)+10=6,356,810.0, and q(5)= 12*10⁷

+Australia (Aus.): 20 pcts. (25 S, 140 E.)

With r(6)= 5,761,227.3 m,  and q(6)= 16*10⁷

(Thunder cloud (negative) makes its opposite patch on land positive (see the super thunderstorm cell), each lightning discharges a lot of charge on both cloud and land charged patch. This is assumed as one of motivation to the surge of E.M every day. Therefore we think about the following problem.)

Figure 6b/II-Land charge (+) contribution-Δ(1)

As matter of fact, the contribution of a positive charge on Earth is contrary to current Earth Magnetism of the Earth.

Problem 1:

A thunder cloud patch of following properties:

-Coverage: 10 km²

-Average negative cloud height: 2000 m

-Location: 30N, 90W as coverage’s centre.

-Total charge: 10⁶ Coulombs.

There is no more charge on cloud left, or the total 10⁶ Coulombs will disappear after the lightning.

Question 1: if we consider the contribution of thunder cloud to E.M, what are the contributions of both cloud and its opposite patch on land to E.M?

Question 2: How is the surge in E.M caused by the lightning?)

This problem is not solved or discussed in this book.

3b-Solution-Ocean contributions (Δ2): We call back I-Biot-Savart.

Δ(i)=7.233869*10^-7*q(i)/r(i)

From ocean facts, the four oceans are covering 70% of global area. Every ocean is connected to the others, the water in ocean is salted and always at high electric conductivity; therefore sea water is considered as homogeneous as a single electrode in relation to any other object.

As assumption in the entry of this problem, our globe is assumed as a neutral object, so the total electric charge is ‘’nil’’. Because the land is charged as counter part of thunder cloud, so the oceans must have charge that opposite to land for neutralizing the land charges. By this argument, we consider the total sea-charge equal to total land-charge and evenly distributed on every ocean and seas connected to them.

Figure 7/II-Globe with Longitude-Latitude network

+From the air-earth current research, our calculation gives an approximate result for total land charge under thunderstorm +80*10⁴*1000 Coulombs. Therefore: 

Total Sea Surface Charge (SSC stands for them from now onward) should be as much as

-80*10⁴*1000 Coulombs or:

=-8*10⁸ Coulombs

We assumed that the total is maintained and distributed evenly over every ocean throughout our calculation.

+From the figure 7/II above, we divide the globe into 6 global chunks (horizontal bands from North down to South). Each chunk expands 30⁰ latitude, we calculate ‘’r’’ (distance to Earth’s rotation axis) from average latitude of the band such as 15 is average of 0-30; 45 is average of 30-60; 75 is average of 60-90.

 After the distance ‘’r’’ calculation, each band can be considered as a rim or a hoop by which the negative charge locates on. The figures are approximately assumed as following:

3/2/1-From 60N to North Pole including Artic, 20*10⁶ km²

Figure 8/II-Arctic and its substitute

-Distance to axis r(1)=Re*cos(75)=1,645,216 m

-Water surface area: 19,889,355 km²

Figure 9/II-Band of 30N-60N latitude degrees and its substitute.

-Distance r(2)=Re*Cos(45)=4,494,936.4 m

-Water surface area: 44,745,958 km²

Figure 10/II- Band of 0-30N latitude degrees and its substitute

Distance to rotation axis: r(3)=Re*Cos(15)=6,140,197 m

Water surface area: 74,576,252 km²

Figure 11/II-Band of 0-30S latitude degrees and its substitute

Distance to rotation axis: r(4)=r(3)=Re*Cos(15)=6,140,197 m

Water surface area: 119,322,568 km2

Figure 12/II- Band of 30-60S latitude degrees and its substitute

Distance to rotation axis: r(5)=r(2)Re*Cos(45)= 4,494,936.4 m

Water surface area: 80,542,806 km2

Figure 13/II- Antarctic and its substitute

-Distance to axis r(6)=r(1)=Re*cos(75)=1,645,216 m

-Water surface area:  14,915,671 km²

X

X                                             X

We are dividing the globe into 6 chunks or bands, each is far from Earth’s rotation axis at an average distance, denoted as r(i) where ‘’i’’ is varying from 1 to 6. We also attribute approximately a certain sea surface area to each chunk in condition that total charge can balance +8*10⁸ Coulombs of all continent summed up.

We have calculated Δ(1) the land contribution to E.M, now we do the same to find out sea surface contribution Δ(2) to E.M. Our job at this stage is to figure out how the 6 chunks contribute to E.M.

Solution:

Total area: 352,876,000 km2,

Charge per s.km: -8*10⁸/352,876,000=-2.267085 Coulombs/km²

We have every q(i) for respective chunk as following:

q(1)= -2.267085*19,889,355=-45,090,858 Coulombs

q(2)= -2.267085*44,745,958=-101,442,890 Coulombs

q(3)= -2.267085*74,576,252=-169,070,702 Coulombs

q(4)= -2.267085*119,322,568=-270,514,404 Coulombs

q(5)= -2.267085*80,542,806=-182,597,387 Coulombs

q(6)= -2.267085*14,915,671=-33,815,094  Coulombs

We again do apply I-Biot-Savart or Biot-Savart 9.1.20 to each chunk from North to South:

Δ(i)=7.233869*10^-7*q(i)/r(i):

-With r(1)=1,645,216 m, q(1)=-45,090,858 Coulombs

 Δ(w1)=-1.982605*10^-5

-With r(2)=4,494,936 m , q(2)=-101,442,890 Coulombs

Δ(w2)=-1.632558*10^-5

-With r(3)=6,140,197 m , q(3)=-169,070,702 Coulombs

Δ(w3)=-1.99185*10^-5 

-With r(4)= 6,140,197 m, q(4) =-270,514,404 Coulombs

Δ(w4)=-3.1869755*10^-5 

-With r(5)=4,494,936 m, q(5)= =-182,597,387 Coulombs 

Δ(w5)=-2.938608*10^-5 

-With r(6)=1,645,216 m, q(6)= =-33,815,094 Coulombs

 Δ(w6)=-1.48682*10^-5    

10.56539*10^-5 Tesla is equivalent to a magnetic composition induced by a charge of 9,315,354.9532*10⁷ Coulomb at Earth equator.

And -13.219416*10^-5 Tesla is equivalent to a magnetic composition induced by a charge of -11,655,372.145666*10⁷ Coulombs at Earth equator.

Figure 14/II-Sea Surface contribution to E.M

Contribution from a negative charge on the Earth is always co-adapted with present Earth Magnetism.

4-Auto-offset Mechanism of the Earth Magnetism:

The aforementioned problems point out that the more positive charge on Earth surface, the more power is taken away from Earth Magnetism. Meantime, for the same speed of a charge crossing perpendicular at the field lines of Earth Magnetism; the more powerful Earth Magnetism, the more powerful force from Earth Magnetism applies on the charge.

Figure 14a/II-Opposite magnetic forces on opposite charged ions

The above mentioned mechanisms lead to a speculation: with the magnetic forces applied on the free ions in the clouds on sky, the Earth Magnetism can offset its magnitude.

That speculation needs to be verified with more test and considerations.

CONCLUSIONS (for part I):

-In addition to underground rotors, the land charges (figure 6b/II) and sea surface charge (SSC), (figure 14/II) contribute to the E.M. Although the land charges as well as sea surface charge are belonging to the Earth but totally influenced by external variableness regardless how we denote it.

-Although those two problems are set and solved with assumed figures, the results are demonstrating that:

+The total of Δ(1)+Δ(2) is on favour of Δ(2), indicating that the influence of the sea is more than that of the land, and supporting E.M or adding on it.

Figure 14b/II-The equivalent of sea&land charges/24 hrs.

+The lightning producing jolt: lightning frees the charges for both thunder cloud and its opposite patch on land; so as consequence of lightning, the Δ(1) must drop right away and considerably. The drop of Δ1 produces a huge magnetic jolt (total Δ(1)+Δ(2) soars due to Δ(1) drop) or the magnetic surge, demonstrated in daily E.M record. Although the lightning is viewed as motivation of every jolt or jerk in Earth Magnetism, but due to the retardation as well as the contribution of many other magnetic inducers, we can point out no direct involvement of lightning in any Earth Magnetism record.

-The divided Earth: might be divided by many different ways, the sum of Δ(1)+Δ(2) might be changed because of such different dividing; but the ultimate demonstration is unchanged and that: The negative charge on sea surface stands up to the positive charge on land surface on term of influence to E.M.

-The solutions to those above problems are leading to a direct relation between Sun position and E.M intensity and E.M 24-hr cycle, but many other influences are piled on E.M and making 24-hr cycle faded. Nonetheless, such cycle will be in another discussion at the end of this book with Maxwell Equation. 

The last chapter presented a discussion about positively charged land surface and the negatively charged sea surface. They both are parts that constituted the Earth and moving together with Earth’s rotation; if the charge of each is opposite to the other, then each induces magnetism that contrarian to the other. In the paragraph 2 (Argument of electric paradox) we even discussed and mentioned the average ratio +ions/-ions in every CC of air on land.

However, we leave that ion ratio for another discussion and work with graph of thunderstorm coverage for our problem in this chapter. So we set and solve the following problems.