The Unification Of Gravity And Magnetism .
The force due to magnetism is the same force we call gravitation.
Justification : A French physicist ( André- Marie Ampère ) noticed the fact that, current carrying wires exhibit a force due to the direction and amount of charge the wires were carrying per unit time .
He realized that when the current in the wires were going the same direction , the wires attracted each other and also repelled each other when going in opposite directions . The photo below illuminates his setting .
F is the force due to magnetism ,
I1 and I2 are the currents carried by wire 1 and wire 2 respectively ,
d is the separation between the wires
µ0 is the permeability of free space( magnetic constant )
l is the length of the wires individually .
For a current of one Ampere going through each wire , the force due to magnetism between the wires separated by a distance of one meter is 2 * 10-7 N/m ie considering a meter long of wires.
A good question will be , how is this related to gravitation ?
It turns out that this same magnitude of force can be computed using different parameters for the same setting ie rather than using the magnetic constant, the Gravitational constant can be employed ; where the charge due to current becomes the number and mass of electrons :
Fg = 2GEML/D
Where ‘
Fg is the force due to gravitation,
G is the gravitational constant(6.6738 *10-11 m3kg-1s-2 ),
E is the number of electrons = (coulomb) 2
= (6.242 * 1018)2
M is the mass of electrons’ factor = m2*Atomic mass unit number,
m is the rest mass of electron = 9.1* 10-31 kg ,
Atomic mass unit number = 1/ Atomic mass unit
= 6.022 * 1026
L is length of wire = 1 meter
D is the separation between wires = 1 meter .
The Atomic mass unit number serves to set the electron rest mass in atomic settings since the electron carrying the charge is actually not at rest ; and notice the fact that it’s factually one thousand times(1000x) the Avogadro number that comprises a mole of entities. A free mind will now see lucidly the relationship between mass and charge.
A computation with the above data says,
F g/m = 2.6 *10-6 N
=4piFm/m
=> Fg/Fm = 4pi .
Where ,
Fm is the force due to magnetismof the charge(one coulomb per second) carrying electrons and has as magnitude 2 * 10-7N per meter of wire.
The justification will not be complete without a first principle approach : A mathematical derivation of the formula FG = 2GEML/D is inevitable .
A good point to start is from the Newtonian gravitation which states that ,
Fg =Gmm2r-2 .
Where r is the distance between m and m2 and G the gravitational constant .
Note should be taken that , we want only the gravitation due to the charge carrying electrons , so we use the amount of current as the argument.
One Ampere means one coulomb of charge per second through a point in space , and one coulomb is 6.242x10^18 charge carriers which in this case are electrons : each wire experiences 6.242x10^18 electron through a point.
Since there are two wires , the total number will be (6.242x10^18)^2.
Also notice the fact that the given mass of the electron is in the rest state , while the actual electrons carrying the charges are not ; To carter for this we express the mass per atomic setting which is the actual scenario : there is no better way to do this than expressing the mass as per atomic mass unit, which involves multiplying the rest mass by 1000 Avogadro number(1/(1.66x10^-27).
From the Newtonian formula one can rewrite the relationship as
FG = GEM/|r|2
This is the force due to a point element on the wire : lets call it dL for wire one and dL2 for wire two ; to know the total force we need to integrate this force along the desired length of wires and this will render the force equation as
FG = GEM∫∫dLdL2 r^/|r|2
From the above picture lets assume that the wires are running vertically(y-axis) and the separation between the wires is horizontal(x-axis) : the equation can be rewritten as
FG = GEM∫∫ dydy2 r^/|((y –y2)2 +D2)1/2|2
Where y and y2 are differential elements on wires one and two respectively .
Also the unit vector r^ pointing in the direction of r is |r|-1 =[ ((y –y2)2+D2)-1/2] :
FG = GEM∫∫dydy2((y-y2)2+D2)-3/2
Integrating indefinitely with respect to dy2 gives
FG =2 GEMD-1 ∫dy .
Integrating again with respect to dy indefinitely will lead to a divergence which is not what we want ,so it will be wise to integrate from zero to L , since we want force per unit length of wire.
After integration we get ,
FG= 2GEMLD-1 .
That's great