Home | Energy | Nuclear | Electricity | Climate Change | Lighting Control | Contacts | Links |
---|

**NUCLEAR MAGNETIC RESONANCE:**

An isolated spheromak in field free space adopts a single 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, axis aligned and axis opposed. 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 related to the energy difference dE between the two states "a" and "b" by the formula:

(Eb - Ea) = dE

= 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 within the particle's spheromak 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 Bz. 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.

Consider a spheromak with central electromagnetic energy density Uo.
from the webpage SPHEROMAK ENERGY the total spheromak electromagnetic energy is:

Ett = [Uo Ro^3 Pi^2 / (2 K)] {4 So [ So^2 - So + 1] / [(So^2 + 1)^2]}

~ 0.96 [Uo Ro^3 Pi^2 / (2 K)]

In the spheromak for R < Rc and extending out to Z = + / - infinity is a region where the magnetic field is axial and the magnetic energy per unit volume is:

[B(R, Z)]^2 / 2 Muo. If an external magnetic field dB is applied and the spheromak rotates between the axis of symmetry aligned and non-aligned positions the energy density change at point (R, Z) is:

dU = [(B + dB)^2 - (B - dB)^2] / 2 Muo

= [( B^2 + 2 B dB + dB^2) - (B^2 - 2 B dB + dB^2)] / 2 Muo

= 4 B dB / 2 Muo

Thus the change in spheromak energy state caused by changing from the aligned to non-aligned states in an externally applied magnetic field dB is:

Integral from Z = - infinity to Z = infinity
Integral from R = 0 to R = Rc of:

dU = (4 B(R, Z) dB / 2 Muo) 2 Pi R dR dZ

As shown on the web page titled: ELECTROMAGNETIC SPHEROMAK in the core of the spheromak the magnetic field energy density as a function of R and Z is given by:

Um = Uo {(Ro^2) / [(K Ro - R)^2 + Z^2]}^2

where:

Um(R, Z) = [B(R, Z)]^2 / 2 Muo

Note that Ro for a proton is much smaller than Ro for an electron and Uo for a proton is much larger than Uo for an electron.

Hence:

B(R, Z) = [2 Muo Um]^0.5

= [2 Muo Uo]^0.5 {Ro^2 / [(K Ro - R)^2 + Z^2]}^2

Thus the change in spheromak energy due to a 180 degree axis alignment rotation in an externally applied magnetic field is:

dE = Integral from Z = - infinity to Z = infinity

Integral from R = 0 to R = Rc of:

[4 dB / 2 Muo] 2 Pi R dR dZ [2 Muo Uo]^0.5 {Ro^2 / [(K Ro - R)^2 + Z^2]}^2}

= Integral from Z = 0 to Z = infinity

Integral from R = 0 to R = Rc of:

[8 dB / 2 Muo] 2 Pi Ro^2 {[2 Muo Uop]^0.5

{R dR dZ / [(K Ro - R)^2 + Z^2]}

= [8 dB / 2 Muo] 2 Pi Ro^2 {[2 Muo Uo]^0.5

Integral from Z = 0 to Z = infinity

Integral from R = 0 to R = Rc of:

[R dR dZ / [(K Ro - R)^2 + Z^2)}

Recall from the web page ELECTROMAGNETIC SPHEROMAK that:

Uo = [Muo C^2 Qs^2 / (32 Pi^2 Ro^4)]

or

{[2 Muo Uo]^0.5} = [2 Muo^2 C^2 Qs^2 / (32 Pi^2 Ro^4)]^0.5

= [Muo C Qs / 4 Pi Ro^2]

Thus:

dE = [8 dB / 2 Muo] 2 Pi Ro^2 {[2 Muo Uo]^0.5

Integral from Z = 0 to Z = infinity

Integral from R = 0 to R = Rc of:

[R dR dZ / [(K Ro - R)^2 + Z^2]}

= [8 dB / 2 Muo] 2 Pi Ro^2 [Muo C Qs / 4 Pi Ro^2]

Integral from Z = 0 to Z = infinity

Integral from R = 0 to R = Rc of:

{R dR dZ / [(K Ro - R)^2 + Z^2]}

= [2 dB] [C Qs]

Integral from Z = 0 to Z = infinity

Integral from R = 0 to R = Rc of:

{R dR dZ / [(K Ro - R)^2 + Z^2])}

= [2 dB] Ro [C Qs]

Integral from (Z / Ro) = 0 to (Z / Ro) = infinity

Integral from (R / Ro) = 0 to (R / Ro) = (Rc / Ro) of:

CONTINUE AND CHECK
= [2 dB]Ro [C Qs] {INTEGRAL}

where:

{INTEGRAL} =
FIX

Note that the value of {INTEGRAL} is independent of the magnitude of Ro and is of the order of [~ 2].

Thus:

dE = [2 dB Ro C Qs {INTEGRAL}]

or

[dE / dB] = [2 Ro C Qs {INTEGRAL}]

A change in spheromak symmetry axis orientation from aligned to opposite with respect to the applied magnetic field dB direction causes a change in the spheromak potential energy. At one 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.

This absorption and emission of photons at a particular frequency that is proportional to the externally applied magnetic field is known as nuclear magnetic resonance (NMR). NMR is a well known non-distructive spectroscopy technique.

Recall that:

[dE / dB] = [2 Ro C Qs {INTEGRAL}]

However:

dE = h F

which gives:

**F / dB = [(2 Ro C Qs / h) {INTEGRAL}]**

**NUMERICAL EVALUATION OF PROTON MAGNETIC RESONANCE:**

F / dB = [(2 Ro C Qs / h) [{INTEGRAL}]

where:

{INTEGRAL} ~ 2

Rearrange the above equation to get:

Ro {Integral} = h (F / dB) / 2 C Qs

**NUMERICAL DATA:**

Muo = 4 Pi X 10^-7 T^2 m^3 / J

Q = 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 = 3.14159265

The experimentally measured value of [F / dB] for protons in water is:

**(Fp / dB) = (42.5781 X 10^6 Hz / T)**

Units:

coul (m / s) T = kg m / s^2

or

coul / kg = (1 / T-s)

Ro {Integral} = h (F / dB) / 2 C Qs

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 OF Ro AND Ett:**

Recall that:

Ro {Integral} = h (F / dB) / 2 C Qs

= [6.62606597 X 10^-34 m^2 kg / s X (42.5781 X 10^6 Hz / T)]

/ [2 X 2.99792458 X 10^8 m / s X 1.602 X 10^-19 coul]

= 29.37168 X 10^-17 m^2 kg Hz s/ s T m coul

= 29.37168 X 10^-17 m (kg / s T coul)

= **29.37168 X 10^-17 m**

Recall that the electromagnetic energy content of a spheromak is given by:

Ett = 0.96 [Uo Ro^3 Pi^2 / (A^2 B)]

where the spheromak's central electromagnetic energy density Uo is given by:

Uo = [Muo C^2 Qs^2 / (32 Pi^2 Ro^4)]

Thus:

Ett = 0.96 [Uo Ro^3 Pi^2 / (A^2 B)]

= 0.96 [Ro^3 Pi^2 / (A^2 B)][Muo C^2 Qs^2 / (32 Pi^2 Ro^4)]

= 0.96 [Muo C^2 Qs^2 / (32 Ro A^2 B)]

The spheromak's contribution to the particle rest mass is given by:

Ett / C^2 = 0.96 [Muo Qs^2 / (32 Ro A^2 B)]

For a proton:

Ro {INTEGRAL} = 29.37168 X 10^-17 m

or

Ro = (29.37168 X 10^-17 m) / {INTEGRAL}

Hence:

Ett / C^2 = 0.96 [Muo Qs^2 / (32 Ro A^2 B)]

= [0.96 X 4 Pi X 10^-7 T^2 m^3 / J X (1.602 X 10^-19 coul)^2][{INTEGRAL} / A^2 B]

/ [32 X (29.37168 X 10^-17 m)]

= 0.032940 X 10^-28 [{INTEGRAL} / A^2 B] T^2 m^3 coul^2 / J m

= 0.032940 X 10^-28 [{INTEGRAL} / A^2 B] Kg^2 m^2 coul^2 / J coul^2 s^2

= 0.032940 X 10^-28 [{INTEGRAL} / A^2 B] Kg

which is a tiny fraction of the proton rest mass of 9.1 X 10^-27 kg. Hence the proton rest mass fraction contributed by the proton spheromak electric and magnetic fields is almost negligibly small.

**NMR CHALLENGES:**

A NMR frequency may be slightly affected by local magnetic fields due to proximity or motion of other nearby charged particles.

The magnetic field Bz experienced by atomic particles is slightly affected by chemical bonding which causes local magnetic field changes. The magnetic field Bz within a sample is also affected by free electrons which move to partially cancel the applied magnetic field. The magnetic field seen by the atomic 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.

In NMR work it is helpful to have a known calibration line such as protons to correct for the effective reduction in the applied magnetic field caused by electrons.

A problem with the NMR technique is that protons and neutrons in nuclei tend to adopt opposing orientations from their nearest neighbours. Hence with an even number of such nucleons there is relatively little change in potential energy on application of an external magnetic field. Thus NMR works best when there is an uneven number of nucleons in a nucleus as with H-1, He-3, Li-7, etc.

**NMR IN MEDICINE:**

Today NMR is widely used in medical imaging where it is used to create images based on the variable spacial concentration of water containing H-1 (protons) within a tissue sample.

**ELECTRON SPIN RESONANCE:**

There is an analogous technique known as electron spin resonance (ESR). However, it is not very useful because in many substances the electrons tend to pair so as to cancel out their spin related magnetic fields.

This web page last updated November 22, 2020.

Home | Energy Physics | Nuclear Power | Electricity | Climate Change | Lighting Control | Contacts | Links |
---|