University of Connecticut OpenCommons@UConn

Chemistry Education Materials Department of Chemistry

June 2006 Invitation to Magnetism and Magentic Resonance Carl W. David University of Connecticut, [email protected]

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Recommended Citation David, Carl W., "Invitation to Magnetism and Magentic Resonance" (2006). Chemistry Education Materials. 9. https://opencommons.uconn.edu/chem_educ/9 Invitation to Magnetism and Magnetic Resonance

C. W. David Department of Chemistry University of Connecticut Storrs, Connecticut 06269-3060 (Dated: June 15, 2006)

I. SYNOPSIS The electric field associated with a charge (Q) at the origin, is With NMR so ubiquitous in modern life, it is reason- able to study spins in magnetic fields, hence this first 1 Q(Coulomb) E = 2 2 attempt. 4π0 r (m )

where we are using a unit test charge (1 Coulomb) to II. UNITS probe the force at the point (x, y, z). It is the magnetic field which causes the problems. Coulomb’s Law First, we start with the Lorentz force: QQ0 F = k ~ ~ ~ r2 F = q~v ⊗ B + qE which includes the electric field just discussed [1]. In the is an empirical law linking the force (F) between two cgs system, B is measured in maxwells/(square cm) i.e, charges, Q and Q’, separated by a distance of r. The in gauss. Then constant is required to make the equation true depending on the units involved. In elementary physics, using the weber maxwell 1gauss = 10−4 = 1 mks system, we have m2 cm2 1 Q(Coulomb)Q0(Coulomb) F (newtons) = (See Figure 2). 4π r2(m2) 0 We start with rationalized units versus unrationalized, where 0 is called the permittivity constant. Its measured see Figure 1. value is about 8.85 × 10−12coul2/(nt − m2). In our atomic/molecular work, it is more convenient to use units which are less common. Specifically, in modified cgs units, we have III. EQUIVALENCE OF ORBITING ELECTRON AND MAGNETIC MOMENT q(statcoulomb)q0(statcoulomb) F (dynes) = r2(cm2) For a magnetic field (vector) B~ acting on an arm of a current loop, a square current loop and a Bohr orbit are which defines a statcoulomb. Thus, a force of 1 dyne is similar (see Figure 3). The force on each single charge exerted by two charges of a statcoulomb each if separated (q) traveling in the arm comes from the Lorentz force [2]: by 1 cm. These are not SI units! But they allow us to do Bohr theory (and other work) without worrying about F~ = q~v ⊗ B~ (3.1) the permittivity of free space! In these units, the charge on the electron is 4.8 × 10−10statcoulomb which is the same as 1.6 × 10−19Coulomb, a convenient reference for Since ~v is perpendicular to B~ (in our case), the cross conversion between the two systems. product simplifies. The current ı is given by:

charge statcoulomb cm statcoulomb ı → n( ) × q( ) × v( ) × σ(cm2) = (3.2) cm3 charge sec sec

where σ is the cross sectional area of the wire-loop, and the force on each charge is:

| F~ |= q | ~v || B~ | sin 90o = qvB (3.3)

Typeset by REVTEX 2 which, in the c.g.s. system is which is

 cm   dyne sec  statcoulomb × = dyne (3.4) dyne sec sec statcoulomb cm 1 T esla = 0.3335683 × 10−6 (3.11) statcoulomb cm where we have defined B’s c.g.s. units as dyne sec and remember that from Coulomb’s law, a statcoulomb (3.5) is a dyne1/2cm. i.e., statcoulomb cm or, in m.k.s. terms, we have for the force (equivalent of dyne sec dyne1/2sec Equation 3.4) 1 T esla = 0.3335683 × 10−6 = dyne1/2cm2 cm2  meter   Newton second  (3.12) Coulomb × × = Newton second Coulomb meter which is the useful c.g.s form of the Tesla (3.6) Since the number of charges is n × a × σ, the force on which defines one Tesla [3] as (the equivalent of Equation one arm of the loop (of length ‘a’) is 3.5)

Newton second 1 T esla = (3.7) F = (n × a × σ) × q × v × B → ıaB (3.13) Coulomb meter or

Newton second 5 dynes −2 meter B ≡ ×10 ×10 (3.8) Coulomb  Newton second  Coulomb meter Newton cm × meter × = Newton or second Coulomb meter (3.14) dyne sec which is reversed on the other (opposite arm) leg of the 1 T esla = 103 (3.9) Coulomb cm loop. and since 1 C = 2.99790 × 109statcoulomb, we have The loop is a×b in area (and 2a+2b in circumference), b the moment arm about the pivot point is 2 sin α if α is dyne sec 1Coulomb the angle between the loop and the field. The moment 1 T esla = 103 × Coulomb cm 2.99790 × 109 statcoulomb arm is of length ‘b/2’, i.e., from the axis to a horizontal (3.10) arm is b/2 (cm). The torque (τ) then is

 b  τ(orque) = |~r ⊗ F~ | = 2 sin α ıaB → Newton meter (3.15) 2

but, since a ×b is the area (A) of the loop, we have Commonly, this torque is related to a magnetic moment equivalent, i.e., τ(orque) = iAB sin α → Newton meter (3.16)

meter2Coulomb  Newton second  ~τ(orque) = ~µ × B~ → → Newton meter (3.17) second Coulomb meter

in analogy with an electric dipole in an electric field. A IV. BOHR THEORY INTERPRETATION OF current loop is equivalent to a magnetic moment, a tiny THE ORBITING ELECTRON’S MAGNETIC bar magnet. MOMENT We assume that the above would hold for a Bohr orbit. At the atomic level, the (Newton) Meter-Kilogram- Second (m.k.s) Scale is a bit cumbersome, especially 3 when we are dealing with charged particles. It is more since convenient to work in the cgs system, with charges in statcoulomb. The major advantage of using statcoulombs q × q0 statcoulomb2 is that it translates directly, without permitivities and F = → = dynes (4.2) dielectric constant considerations, into forces when em- r2 cm2 ployed in Coulomb’s Law. This allows us to write that

1 statcoulomb = 1 dyne1/2cm (4.1) From Bohr Theory we had

n2¯h2 erg2 sec2  erg2 sec2  r = → = = cm (4.3) Zme2 (gram statcoulomb2) (gram dyne cm2)

and [4] so, solving for θ˙ we have

2 ˙ mr θ = n¯h = pθ → erg sec (4.4)

n¯h erg sec dyne sec gm (cm/sec2) sec 1 θ˙ = → = = = (4.5) mr2 gram cm2 g cm g cm sec

and then the current (= statcoulomb/sec) = so charge/transit-time = −e/τ, where [5] τ is the pe- riod. ∆θ n¯h 2π = θ˙ = = (4.6) ∆t mr2 τ

charge e en¯h statcoulomb erg sec statcoulomb = = i = − → = (4.7) period τ 2πmr2 gram cm2 sec but, since the area is πr2, we have en¯h en¯h statcoulomb cm2 statcoulomb erg sec iA = − πr2 = − → = (4.8) 2πmr2 2m sec gram which defines the Bohr magnetic moment

e¯h dyne1/2cm erg sec dyne1/2cm2 µ = − → = (4.9) B 2m gram sec i.e.,

4.8 × 10−10statcoulomb × 6.627 × 10−27erg sec dyne1/2cm2 = = 0.27781 × 10−9 (4.10) 2 × 9.1094 × 10−28grams × 2 × π sec known as the Bohr magneton. This works out to be (since the charge on the electron is 1.6 × 10−19Coulomb and 4.8 × 10−10statcoulomb)

 −19  103 grams −9 statcoulomb erg second 1.6 × 10 Coulomb kilograms 0.27788 × 10 −10 × 7 erg (4.11) T esla 4.8 × 10 statcoulomb 10 Joule 4

Joule Coulomb second = 0.0926 × 10−22 (4.12) kilogram

Joule Coulomb second Newton second J = 0.0926 × 10−22 Coulomb meter = (4.13) kilogram T esla T which compares to the literature values of 9.27402 × 10−24J/T . or 9.27402 × 10−21erg/G.

V. THE NUCLEAR MAGNETIC MOMENT HOWEVER, the true (measured) nuclear magnetic moment (for a bare proton) is µproton = 2.79277µN . The nuclear magnetic moment, obtained by changing the mass from that of the electron to that of the proton, is J 9.1094 × 10−28 VI. THE TORQUE µ = 9.26 × 10−24 × (5.1) N T 1.67 × 10−24 which is, in mks, We then have

−28 J en¯h 50.5 × 10 (5.2) τ(orque) = ıAB sin α = − B sin α (6.1) T 2m which compares with literature values of 5.0505 × 10−24erg/G i.e., 5.0508 × 10−27J/T which is, finally (using Equation 4.9),

dyne1/2cm2   dyne sec  τ(orque) = nµ B sin α = ~µ × B~ → × = dyne cm (6.2) B sec statcoulomb cm

The magnetic moment, ~µ, is proportional to the an- which gives gular momentum (` = n¯h, Bohr), with a proportionality constant γ, i.e., E = −nµBB cos α (7.3)

e¯h or ~µB = = γ¯h (6.3) 2me E = −n~µB · B~ → dyne cm = ergs (7.4) so as the angular momentum precesses in an external magnetic field, so does the magnetic moment vector, vide infra. VIII. LARMOUR’S THEOREM

We have VII. ENERGY OF AN ORBITING ELECTRON IN AN ARBITRARY MAGNETIC FIELD m~r¨ = F~ + e~r˙ × B~ (8.1)

The energy associated with rotating the current loop If we specialize (arbitrarily) to make the magnetic field (Bohr orbit) about the axis perpendicular to the field point along the z-axis, then (orienting the loop relative to the field), is mx¨ = Fx + eBy˙ (8.2) Z α, final angular position E = τ(α)dα (7.1) reference position my¨ = Fy − eBx˙ (8.3) i.e., and Z α E = τ(x)dx (7.2) 90o mz¨ = Fz (8.4) 5

What is F~ ? It is (see Figure 4). This set of equations are used to transform from Ze2rˆ statcoulomb2 F~ = = → dynes (8.5) {x(t),y(t)} ,→ {x’(t),y’(t)}. The reverse of this can be r3 cm2 obtained in two steps. First, multiply the top equation i.e., the Coulomb force. Z is the atomic number, e is the (1) by cosine, and the lower one (2) by sine, and add. We charge on the electron (in dyne1/2cm, i.e., statcoulombs). obtain This means x(t) = x0(t) cos ωt + y0(t) sin ωt (8.12) Ze2x F = (8.6) x 3 and then multiple the top equation by sine and the lower px2 + y2 + z2 one by cosine, and subtract, yielding with similar terms for y and z, y(t) = −x0(t) sin ωt + y0(t) cos ωt (8.13) Ze2y Fy = 3 (8.7) which, in matrix form is: px2 + y2 + z2  x(t)   cos ωt sin ωt   x0(t)  = 0 (8.14) Ze2z y(t) − sin ωt cos ωt y (t) F = (8.8) You tell me an instantaneous value of x and y, and I will z p 3 x2 + y2 + z2 tell you the instantaneous values of x’ and y’ (Equation 8.11), and vice versa using Equation 8.14. It is common to begin treating Larmour precession us- It is interesting to substitute the result of one trans- ing a rotating co¨ordinate system attached to the original formation into the other. This should result in ‘no trans- one, sharing a common z-axis. Therefore, we need to formation’, since we are doing and undoing the transfor- make a ”side trip” to co¨ordinate transformations. The mation. We have: situation at hand concerns two dimensions, x and y. Rel- ative to this ‘couple’ {x(t),y(t)}, we wish to introduce  x(t)   cos ωt sin ωt   cos ωt − sin ωt   x(t)  = one which is rotating i.e., {x’(t),y’(t)}, where the primed y(t) − sin ωt cos ωt sin ωt cos ωt y(t) co¨ordinate system is rotating about the z-axis relative to the original {x(t),y(t)} system. Once per revolution, x’ which is and y’ will lie exactly juxtaposed upon x and y. The transformation equations are:  cos ωt sin ωt   cos ωt − sin ωt   1 0  = − sin ωt cos ωt sin ωt cos ωt 0 1 x0(t) = x(t) cos ωt − y(t) sin ωt (8.9) which seems perfectly reasonable! y0(t) = x(t) sin ωt + y(t) cos ωt (8.10) which we re-write in matrix notation A. Taking time derivatives of co¨ordinates  x0(t)   cos ωt − sin ωt   x(t)  = (8.11) It is clear that y0(t) sin ωt cos ωt y(t)

dx(t) dx0(t) dy0(t) x˙ = = cos ωt + sin ωt − x0(t)ω sin ωt + y0(t)ω cos ωt (8.15) dt dt dt and dy(t) dx0(t) dy0(t) y˙ = = − sin ωt + cos ωt − x0(t)ω cos ωt − y0(t)ω sin ωt (8.16) dt dt dt and therefore d2x(t) x¨ = =x ¨0(t) cos ωt +y ¨0(t) sin ωt + 2ω (−x˙ 0 sin ωt +y ˙0 cos ωt) − ω2 (x0(t) cos ωt + y0(t) sin ωt) dt2 so d2y(t) y¨ = = −x¨0(t) sin ωt +y ¨0(t) cos ωt + 2ω (−x˙ 0 sin ωt +y ˙0 cos ωt) + ω2 (−x0(t) sin ωt + y0(t) cos ωt) dt2 6 and substituting into Equation 8.2

x¨ =x ¨0(t) cos ωt +y ¨0(t) sin ωt + 2ω (−x˙ 0 sin ωt +y ˙0 cos ωt) − ω2 (x0(t) cos ωt + y0(t) sin ωt)

F eB = x + (−x˙ 0 sin ωt +y ˙0 cos ωt − ω(x0(t) cos ωt + y0(t) sin ωt)) m m and,

y¨0 = −x¨0(t) sin ωt +y ¨0(t) cos ωt + 2ω (−x˙ 0 cos ωt +y ˙0 sin ωt) + ω2 (−x0(t) cos ωt + y0(t) sin ωt) so, substituting into Equation 8.3 F eB y¨0 = y + (−x˙ 0 cos ωt +y ˙0 sin ωt − x0(t)ω sin ωt + y0(t)ω cos ωt) m m

We note that the Coulomb force on the electron is not ef- externally, we see a magnetic dipole in space, interacting fected substantially by the transformed co¨ordinates, since with external (to it) magnetic fields. How does it react? x and y are merely ”changing places”, while r stays con- We start with the elementary definition of torque, stant (z stays constant no matter what) (see Figure 5). First, we assume that terms of the order ω2 are small L~ = ~r ⊗ ~p enough to neglect. Second, we start our clock at t=0 exactly when the x(t) and x’(t) axis are superimposed. so that Then, we have: dL~ ~p x¨0(t) cos ωt + 2ω (y ˙0 cos ωt) − ω2 (x0(t) cos ωt) = ⊗ ~p + ~r ⊗ F~ = ~r ⊗ F~ = ~τorque dt m

F eB where we have assumed Newton’s second Law in equate = x + (y ˙0 cos ωt + x0(t)ω cos ωt) (8.17) m m the time derivative of the momentum with the force, and that ~p ⊗ ~p is zero! and ~τ(orque) = ~µ × B~ (9.1) y¨0(t) cos ωt + 2ω (y ˙0 cos ωt) − ω2 (y0(t) cos ωt) ~ d~µ ~ ~ and ~µ = γL Since dt = −γB ⊗ µ (and we assume the B Fy eB points along the z-axis) which is = + (y ˙0(t) cos ωt + x0(t)ω cos ωt) (8.18) m m

ˆi ˆj kˆ (see Figure 6). If we were to choose dµx dµy dµz ˆi + ˆj + kˆ = −γ 0 0 B dt dt dt eB statcoulomb dynesec µx µy µz 2ω = → statcoulomb cm = sec−1 m gram we see immediately that the z-component of the magnetic (see Equation 3.4) then the velocity terms would disap- moment never changes! [6] Expanding, we have pear! This is the Larmour frequency: dµ x + γBµ = 0 eB y ω = dt 2m Approximately, if we choose to rotate a co¨ordinate sys- dµ y + γBµ = 0 tem at this particular frequency, then the magnetic mo- dt x ment would appear stationary in this rotating co¨ordinate system. a solution of which is

~µ = (AcosγBt)ˆi + (AsinγBt) ˆj + a constant kˆ IX. PRECESSION OF A MAGNETIC MOMENT which means that the µ vector is precessing about the z- Regardless of whether a magnetic moment originates axis, with its x- and y-components interchanging during in orbital motion of charges or intrinsically in a nucleus, the precession. 7

X. AN ALTERNATIVE DERIVATION OF THE XI. NMR PRECESSION If we assume that the magnetic moment we are dealing with is due to a proton, i.e., a spin one half particle, then we know there are two orientations possible, i.e., ~τorque = γ~µ ⊗ B~ two possibilities of the state of the proton, one with the magnetic moment ‘up’, and the other ‘down’. The energy is the torque on a spin (1/2) particle in a static magnetic of these two states are field B~ . Then 1 E = γB d~µ down 2 = ~τorque dt and or 1 E = − γB d~µ up = −γB~ ⊗ ~µ 2 dt with a ∆E If we write ∆E = γB = g µ B   N N ~µ = −|~µ| ˆi sin θ cos φ + ˆj sin θ sin φ + kˆ cos θ (see Figure 7). where gN = µ is the “nuclear factor”, and B is the magnetic field strength (see Figure 8). Empiri- then cally, gN = γ = 5.5856912. One has, then, ˆ ˆ ˆ i j k ∆E g µ B B~ · ~µ = −|~µ| 0 0 B = ν = N N sin θ cos φ sin θ sin φ cos φ h h which, for a 1 Tesla Field would be So −27 5.5856912 × 5.0508014 × 10 J/T × 1T 7 −1 dµx d|~µ| sin θ cos φ ν = = 4.26×10 seconds = 42.6MHz = −γB|~µ| sin θ sin φ = (10.1) 6.627 × 10−34J − sec dt dt dµ d|~µ| sin θ sin φ y = γB|~µ| sin θ cos φ = (10.2) dt dt XII. PERTURBING WITH AN EXTERNAL dµ d|~µ| cos θ z = 0 = (10.3) MAGNETIC FIELD dt dt In the Continuous Wave (CW) experiment, we super- The last equation may be integrated directly, leaving impose a rotating magnetic perturbation field on the above system, choosing to apply that field in the x-y cos θ = constant plane only, at right angles to the main field which orig- inally set up the energy differences. We make this field assuming the |~µ| is constant (as it must be). B cos ωt where the strength of the perturbation B is In order to integrate Equation 10.1 and Equation 10.2 o o small, and the frequency ω is tunable. As we change we write ω we tune through the frequency ω set up by the main d sin θ cos φ dφ magnetic field, and at this ωtuned value, transitions are γB sin θ sin φ = = − sin θ cos φ (10.4) dt dt observed. d sin θ sin φ dφ −γB sin θ cos φ = = sin θ sin φ (10.5) dt dt XIII. BLOCH’S EQUATIONS which means that In order to talk about spin relaxation, one needs dφ −γB = to first obtain the differential equations governing the dt macroscopic spin. Normally, this discussion concerns the or macroscopic magnetic moment, rather than individual spins. The macroscopic magnetic moment is normally φ = −γBt + C called M~ . with three components, one of which, cho- sen arbitrarily, coincides with the z-axis and the external where C is a constant (usually called the phase angle). applied magnetic field. 8

If the microscopic spins have aligned themselves prop- XIV. ALTERNATIVE VIEW OF CURRENT erly in the external field, with Nα in the α state and Nβ LOOP (ADDRESSING SPIN-SPIN ETC., in the β state, then COUPLING ORIGINS

Mz = γN ¯h(Nα − Nβ) We can view the current loop as a single turn of a solenoid, i.e., creating a magnetic field which passes where γ is the magnetogyric ratio appropriate. N through the center of the loop. We have, using the Law We know that at thermal equilibrium of Biot and Savart

Nα g eh¯ B/kT g µ B/kT = e N 2mc = e N B dσ~ × ~r Nβ δB~ = i r3 When the field is turned off, the z-component of mag- netic moment will approach its final value exponentially dσ × r cos 90o dσ in time, so that the above ratio approaches 1. δB~ = Ki = Ki r3 r2 1 dM N − N z = − α β where δB~ is perpendicular to the ~r vector, indicated as γN ¯h dt T1 δB and K is a proportionality constant whose value is exactly 10−7 weber which is often written as where T1 is defined as the spin-lattice relaxation charac- ampere meter teristic time. or µo dM M = K z = − z 4π dt T 1 where When the magnetic field is different from zero but con- −7 weber stant, Mz tends to a constant value, M0, related to the µo = 12.57 × 10 macroscopic static magnetic susceptibility. Now, one has ampere meter (see Figure 9). on the figure. The component of δB along dMz Mz − M0 = − the x-axis from this piece of current loop is dt T1 µ dσ and δB = o i sin α x 4π r2 M M x = − x dt T and if we add up the contributions from all elements of 2 the loop by integrating around it, obtaining 2πR (the and circumference) we obtain

My My µ sin α2πR = − B = o i dt T2 x 4π r2 where T2 is the transverse relaxation time. But sin α is R/r, so we obtain Since the bulk magnetic moment is the sum of all the individual magnetic moments, and since they are actually µ πR2 B = o i precessing about the magnetic field, one has x 4π r3 dM~ and if we now translate the point of interest to the origin, = γ M~ ⊗ B~ dt N so that r → R, we obtain in the absence of relaxation. We know that this latter µ πR2 weber ampere weber B = o i → = = tesla leads to x 4π R3 ampere meter meter meter2

Mx Mx = − + γN MyBz dt T2 µo i Newton second Bx = → tesla = and 4 R Coulomb meter This result says that there is a magnetic field at the My My = − + γN MxBz origin, caused by the current loop. It is as if there were dt T 2 no current loop. but instead there was a little bar mag- and finally net at the origin (see Figure 10). It is clear that if the current element is an electron (in an atomic orbit) and if dM M − M z = − z 0 that electron carries a spin with it, then that spin would dt T1 interact with the magnetic field generated by the dipole 9 at the origin (itself the result of the electron orbiting), so where γ is the magnetogyric ratio. We are equating the that there would be two states, one in which the electron magnetic moment and the angular momentum through a spin was ‘up’ relative to the magnetic dipole’s field, and proportionality constant. This means that for a classical the other ‘down’. This is a quasi-classical model for spin orbit with ` 6= 0 there exists an equivalent magnetic mo- orbit interaction, wrong in several details, but sufficient ment, located at the origin (which is where the nucleus to illustrate approximately where spin orbit interaction lies), i.e., a tiny bar magnet associated with the orbital comes from. angular momentum. We had

ne¯h µ = − 2m for any n, and we can write this as XV. FIGURES e µ =| ~µ |⇒ ~µ ≡ L~ ≡ γL~ 2m

cgs (cm,g,sec) absolute

esu emu

qq’ F = unrationalized F in dynes r 2 at q=q’=1, r=1 r in centimeters qq’ F = rationalized F in dynes/4π q in statcoulomb 4π r 2

FIG. 1: Unrationalized versus rationalized units

[1] In order to remember directions in the cross product, one [3] 10−4T esla = 1Gauss. need only remember that [4] remember, e = 4.8 × 10−10statcoulomb, or 1.6 × 10−19Coulomb ˆi ⊗ ˆj → kˆ [5] we use τ twice here, once for the torque, and once for the period- both usages are common [2] If there is an electric field, the Lorentz force would be [6] Reiterating, B~ = Bkˆ (a convenience). F~ = q~v ⊗ B~ + qE~ 10

cgs (cm,g,sec) absolute

esu emu Gaussian mixed Popular practical rat rat unrat Very Popular absolute emu + (m,kg,sec)

unrat rationalized if rectangular symmetry unrationalized if spherical symmetry

FIG. 2: Units, units, units

Axis a F = i a B normal α i τ=( iaB)( b sinα ) τ= iAB sin α A=ab=area Axis B F = i a B b

F Axis

normal

e− τ= eAB sinα A=π r2 Axis

B F

FIG. 3: A Current Loop in a Magnetic Field 11

z

y’ y

x

x’

FIG. 4: Relationship between ‘stationary’ and ‘rotating’ co¨ordinates 12

z B

y

x

y

x

FIG. 5: Precessing Magnetic Moment in Magnetic Field 13

z B

= o µx µx cos ω t

= o µ y µy sin ω t y 0 a 2a

x perturbing field in the x−y plane B’ = B cos o Ω t y

0 2a a

x

FIG. 6: Relation of Precessing Magnetic Moment to Instru- mental Perturbing Magnetic Field in x-y plane

B z z main field

E = (1/2) γ B exciting field B x ’=Bx ’ cosω t y B =B cos t y y’ y’ ω

E = − (1/2) γ B x x

FIG. 7: NMR Transition, from Low to High Energy State 14

B, field strength

Energy ∆ E = hν

FIG. 8: NMR Transition at a Chosen Field Strength

z

dσ

R r δΒcos α δΒ α y δΒsinα x

FIG. 9: A Current Loop Creating a Magnetic Field 15

z

S

N y electron with its spin interacting with orbitally generated x magnetic field

FIG. 10: A Current Loop Creating a Magnetic Field is Equiv- alent to a Bar Magnet