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