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] Follow this and additional works at: https://opencommons.uconn.edu/chem_educ 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).