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NUCLEAR MAGNETIC RESONANCE

By Charles Rhodes, P.Eng., Ph.D.

NUCLEAR MAGNETIC RESONANCE:
An isolated spheromak in field free space adopts a single low energy state. However, if there is an externally applied magnetic field the energy of the otherwise isolated spheromak becomes a function of the spheromak orientation with respect to the externally applied magnetic field. Each spheromak has two extreme values for its orientation dependent energy, aligned and unaligned. Then the spheromak can transition from its lower energy state to its higher energy state by absorption of a photon or can transition from its higher energy state to its lower energy state by emission of a photon. The frequency Fp of the emitted or absorbed photon is approximately related to the energy difference dE between the two states by the formula:
(Ea - Eb) = dE
= h (Fa - Fb)
= h Fp
where:
h = Planck constant.

Thus if an atomic charged particle is located within an externally applied magnetic field the net particle charge Q remains unchanged and the charge circulation velocity C remains unchanged, but the particle's potential energy is dependent on the orientation of the particle's spheromak symmetry axis with respect to the external magnetic field Bx. A change in the spheromak's symmetry axis orientation with respect to the axis of the applied external magnetic field causes a change in the particle's potential energy which change is accompanied by emission or absorption of a photon.

A change in spheromak symmetry axis orientation with respect to the applied magnetic field direction causes the spheromak poloidal loop current to interact with magnetic field Bx which causes a torque on the particle. This torque, acting through the changing angle between the two axes, causes a change in particle potential energy. At one particle orientation extreme the applied magnetic field vectorially reduces the spheromak's poloidal magnetic flux. At the other particle orientation extreme the applied magnetic field vectorially increases the spheromak's poloidal magnetic flux. The transition between these two energy extremes requires rotating the spheromak axis of symmetry through 180 degrees = Pi radians.

If RF energy of the appropriate frequency is applied the charged particles will tend to absorb photons and be excited from their lower energy state to their higher energy state. If the RF energy supply is removed the particles will tend to emit photons and decay from their higher energy state to their lower energy state. The frequency Fp of these photons is dependent via the Planck constant on the energy difference between the two particle energy states that in turn is dependent on the strength of the externally applied magnetic field and is inversely proportional to the spheromak's contained field energy.

This absorption and emission of photons at a particular frequency that is proportional to the applied magnetic field and is proportional to the spheromak magnetic moment M is known as nuclear magnetic resonance (NMR). This frequency may also be slightly affected by local magnetic fields due to proximity or motion of other nearby charged particles.

Note that bulk motion of free electrons reduces the strength of the externally applied magnetic field Bx as seen by the charged particle spheromaks. In NMR work it is helpful to have a known calibration line such as protons to correct for the reduction in the applied magnetic field.

Today NMR is widely used in medical imaging where it is used to creates images based on the variable spacial concentration of water (protons) within a tissue sample.
 

NMR ANALYSIS:
This web page relies on parameters developed on the web page titled: SPHEROMAK MAGNETIC MOMENT.

Let the external field be Bx. A spheromak can be viewed as a constant current loop that interacts with the applied magnetic field. The magnetic moment M of the spheromak can be measured using the formula:
M = (h / 2) (dFh / dBx)
where:
dFh = the frequency of photons absorbed or emitted
dBx = the externally applied magnetic field
h = Planck Constant.

The magnetic field dBx sensed by atomic particles is slightly affected by chemical bonding which causes local magnetic field changes. The magnetic field Bx in the sample is also affected by free electrons which move to partially cancel the applied magnetic field. The magnetic field seen by the particles is also affected by the adjacency of other particles with magnetic moments. Note that the applied magnetic field must be highly uniform through the sample to properly resolve the narrow NMR spectral lines.

Note that bulk motion of free electrons reduces the strength of the externally applied magnetic field Bx as seen by the spheromaks. In NMR work it is helpful to have a known particle species present such as protons to correct for the effect of the reduced applied magnetic field.
 

NUMERICAL DATA:
Mu = 4 Pi X 10^-7 T^2 m^3 / J
Qa = 1.602 X 10^-19 coul
C = 2.99792458 X 10^8 m / s
h = 6.62606597 X 10^-34 m^2 kg / s
Me = 9.109 10-31 kg
Pi = 3.14159
Np = 222
Nt = 305
Nr = 0.7278688525
Nr^2 = 0.5297930664
So = 2.025950275
So^2 = 4.104474517

Units:
coul (m / s) T = kg m / s^2
or
coul / kg = (1 / T-s)
or
T = kg / coul s
and
J = kg m^2 / s^2
giving:
J / T = kg m^2 coul s/ s^2 kg = m^2 coul / s

Thus:
J / T^2 m^2 coul = (m^2 coul) / (s T m^2 coul) = 1 / s T
 

NUMERICAL EVALUATION:
[So^2 / {[Nr^2 (So^2 + 1)^2] + [(So^2 - 1)]^2}]
= [4.104474517 / {[0.5297930664 (5.104474517)^2] + [(3.104474517)]^2}]
= [4.104474517 / {[13.80410806] + [9.637762027]}]
= [4.104474517 / {23.44187009}]
= 0.175091599

[2 Q Nr / Pi Nt Me] = [2 X 1.602 X 10^-19 coul X 0.727869 / (Pi X 305 X 9.109 X 10^-31 kg)]
= 2.67193 X 10^-4 X 10^12 coul / kg
= 2.67193 X 10^8 coul / kg

Thus:
dFh / dBx = [2 Q Nr / Pi Nt Me] [So^2 / {[Nr^2 (So^2 + 1)^2] + [(So^2 - 1)]^2}]
= [2.67193 X 10^8 coul / kg] [0.175091599]
= 0.467833 X 10^8 coul / kg
= 46.7833 X 10^6 coul / kg
= 46.733 X 10^6 (1 / T-s)
= 46.733 MHz / T

This value compares to the experimentally measured value for protons of:
(dF / dBx) = (42.5781 X 10^6 Hz / T)
even though the mass Me of an electron was used as the electromagnetic energy content of the proton spheromak in the above calculation.
 

PHYSICAL SIZE OF A CHARGED PARTICLE SPHEROMAK:
Find the nominal radius Ro for proton spheromaks:

Assume that the total proton spheromak field energy is approximately equal to the rest mass energy of an electron. Then from the web page titled: SPHEROMAK ENERGY
Efs = (Mu C^2 / 2)[Qa / 4]^2 [1 / Ro]
and
Ett = (Ett / Efs) Efs
= Me C^2

Thus:
Efs = Me C^2 / (Ett / Efs)
= (Mu C^2 / 2)[Qa / 4]^2 [1 / Ro]

Hence:
(Mu C^2 / 2)[Qa / 4]^2 [1 / Ro] = Me C^2 (Efs / Ett)
or
Ro = (Mu / 2)[Qa / 4]^2 [1 / Me] [Ett / Efs]

From the web page titled: PLANCK CONSTANT
(Ett / Efs) = {1 - [(So -1)^2 / (So^2 + 1)]^2}

So = 2.025950275
So^2 = 4.104474517

Numerical substitution gives:
(Ett / Efs) = {1 - [(So -1)^2 / (So^2 + 1)]^2}
= {1 - [(1.025950275)^2 / (5.104474517)]^2}
= 0.959602867

Then: Ro = [4 Pi X 10^-7 T^2 m^3 / J][1 / 2][1.602^2 X 10^-38 coul^2 / 16][0.959602867] / 9.109 X 10^-31 kg
= 0.10613 X 10^-14 T^2 m^3 coul^2 / J kg
= 1.0613 X 10^-15 (kg / coul s)^2 m^3 coul^2 / J kg
= 1.0613 X 10^-15 (kg m^3 / s^2 J)
= 1.0613 X 10^-15 m
= 1.0613 fermi
 

This figure compares to a published "proton charge radius" in the range .84 fermi to 0.87 fermi.
 

QUANTIFICATION OF Bpo:
Recall from web page titled: NUCLEAR MAGNETIC FLUX QUANTUM that:
Bpo Ro^2 Pi = Phip / {Ln[1 + (1 / So^2)]}
or
Bpo = [Phip / {Ro^2 Pi Ln[1 + (1 / So^2)]}
= [3.290157699 X 10^-18 T m^2 / {(1.0613 X 10^-15 m)^2 Pi Ln[1 + (1 /4.104474517)]}
= [3.290157699 X 10^-18 T m^2 / {(1.12635769 X 10^-30 m^2) Pi [0.2180397]}
= 4.264 X 10^12 T

Thus Bpo for a charged particle spheromak with electromagnetic field energy equal to an electron rest mass is about 4.264 X 10^12 T

Hence the externally applied magnetic field is tiny compared to the spheromak core magnetic field strength.

An important conclusion from the above work is that assuming that protons are sensed by conventional NMR much of a proton's rest mass is not directly due to its electric and magnetic fields whereas with an electron its rest mass is almost entirely due to its electric and magnetic fields.
 

A problem with the NMR technique is that protons and neutrons in nuclei tend to adopt opposing orientations from their nearest neighbours. With an even number of such nucleons there is relatively little change in potential energy when a nucleus changes orientation in an external magnetic field. Thus NMR works best when there is an uneven number of nucleons as with light hydrogen.
 

ELECTRON SPIN RESONANCE:
There is an analogous technique known as electron spin resonance (ESR). However, it is not very useful because the electrons tend to pair so as to cancel out their spin related magnetic fields. Also, for electrons the spheromak field energy Ett is much smaller than for protons which leads to very high operating frequencies.
 

This web page last updated February 5, 2018.

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