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.