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Let the dimensionful angular momentum operators obey $\mathbf{F}=\mathbf{I}+\mathbf{J}$ and take the interaction Hamiltonian as $H^{\prime }=A\mathbf{I}\cdot \mathbf{J}$. Sincethe energy eigenvalues at fixed $I,J$ arewhere $F=|I-J|,\dots ,I+J$. Subtraction gives the Landé interval ruleMany atomic conventions use dimensionless angular momentum operators and call the energy-valued constant itself $A$; then the right-hand side is $AF$. Each energy level has $2F+1$ magnetic sublevels with the same energy in zero field.

For fine structure, the effective interaction is $a_{LS}\mathbf{L}\cdot \mathbf{S}$ with dimensionless angular momentum operators. It originates in the spin-orbit interaction, including the magnetic interaction of electron spin with orbital motion and its relativistic correction. Within an isolated LS coupling atomic term, the labels are the electron configuration, $L,S,J,M_J$, and the interval between adjacent $J$ energy levels is $a_{LS}J$. This is smaller than the electrostatic separation of atomic terms.

For magnetic-dipole hyperfine structure, the nuclear magnetic dipole moment interacts with the magnetic field generated by the electrons. Contributions include the Fermi contact interaction of penetrating $s$ electrons, orbital magnetic fields, and the dipolar field of electron spin. Projected onto one electronic $J$ energy level, their rotationally invariant magnetic-dipole part has the form $a_{IJ}\mathbf{I}\cdot \mathbf{J}$. The labels are $I,J,F,M_F$, together with the approximate electronic labels such as $L,S$. This splitting is generally smaller than fine structure, reflecting the much smaller nuclear magnetic dipole moment. An electric nuclear quadrupole moment can add a different interaction and violate the simple Landé interval rule.

At the larger electrostatic energy scale, an effective two-spin exchange interaction can also be written $a_s{\mathbf{s}}_1\cdot {\mathbf{s}}_2$ within fixed orbitals. Its origin is the Coulomb interaction combined with the Pauli exclusion principle, rather than a large direct magnetic interaction. It separates singlet state and triplet state atomic terms; before spin-orbit coupling, useful labels include the electron configuration, $L,M_L,S,M_S$. General electrostatic atomic term separations need not obey a single Landé interval rule as $L$ varies.

The specified two-direction constraint permits magnetic moments ${\mu }_z=\pm \mu$, withwhere ${\mu }_B$ is the Bohr magneton. There are four magnetic gadolinium ions per conventional face-centered cubic lattice cell; the closed-shell nitrogen ions add no local spin moment. Hence the magnetic-ion number density isAt high temperature, neglect interactions and use $B_z\simeq {\mu }_0{H}_z$. The one-ion partition function and mean magnetic dipole moment areIn the weak-magnetic field limit, $\tanh x\simeq x$, so $M_z=n⟨{\mu }_z⟩={\mu }_{0}n{\mu }^2{H}_z/(k_{B}T)$. Thus the dimensionless SI magnetic susceptibility is $\chi =C/T$, withThe coefficient has units of kelvin. The unrestricted-spin expression proportional to $S(S+1)/3$ is inappropriate here: the source explicitly retains only the two orientations with $S_z=\pm S$. In this two-state Curie law, the squared axial moment is $(g_{J}S{\mu }_B{)}^2$.

Determining antiferromagnetic magnetic order is a good reason to use neutron scattering. The neutron has a magnetic dipole moment, so its scattering amplitude includes the spatial arrangement of electronic magnetic moments. Alternating magnetic moments can produce magnetic neutron diffraction peaks even when there is no change of the nuclear crystal structure. Ordinary nonresonant X-ray diffraction primarily measures the electron density, and is much less sensitive to this magnetic order.

Let ${\rho }_q=3q/(4\pi a^3)$ be the uniform charge density and take $\mathbf{\omega }=\omega \hat{\mathbf{z}}$. Rigid rotation produces current density $\mathbf{J}={\rho }_q\mathbf{\omega }\times \mathbf{r}$. The magnetic dipole moment isSpherical symmetry gives $\int r_i{r}_j{d}^{3}r={\delta }_{ij}\int r^2{d}^{3}r/3$, while $\int r^2{d}^{3}r=4\pi a^5/5$. ThereforeIt is parallel to the angular velocity for positive $q$ and antiparallel for negative $q$. The fully signed magnitude statement is $|{\mathbf{m}}_s|=|q||\omega |a^2/5$; the printed magnitude assumes positive $q$ and positive angular speed.

Axisymmetry permits a purely azimuthal magnetic vector potential, $\mathbf{A}=f(r)\sin \theta \hat{\mathbf{\phi }}$, with $A_r=A_{\theta }=0$. For the source's associated Legendre function convention, $P_1^1(\cos \theta )=+\sin \theta$, since its definition omits the Condon–Shortley phase. The given differential operator reduces toInside the sphere, $J_{\phi }={\rho }_q\omega r\sin \theta$. Since the radial operator maps $r^3$ to $10r$, the particular coefficient is ${\alpha }_0=-{\mu }_0{\rho }_q\omega /10$. Regularity at the origin and decay at infinity giveThere is no surface sheet electric current, so both normal and tangential magnetic field components are continuous at $r=a$. Usingthis is equivalent to continuity of $f$ and $f^{\prime }$. The two equations give $A={\mu }_0{\rho }_q\omega a^2/6$ and $C={\mu }_0{\rho }_q\omega a^5/15$. ThusIn the homogeneous expansion, ${\alpha }_1=A$, ${\beta }_1=C$ in their respective interior and exterior regions; all higher angular orders vanish. Indeed, matching the regular $r^{\ell }$ and decaying $r^{-\ell -1}$ terms with zero source at $\ell >1$ forces both coefficients to zero. The outside magnetic field is exactly the magnetic dipole field of ${\mathbf{m}}_s$, not merely its leading far-field approximation. For a positive moment along $z$, $A_{\phi }={\mu }_0{m}_s\sin \theta /(4\pi r^2)$, which is the printed cross-product magnitude expression.

For the flux coupling of a rotating charged sphere to a circular loop, take a loop of radius $b>a$ in the equatorial plane and use Stokes theorem on the regular full vector potential, evaluated around the rim. Its linked magnetic flux isWith $m_s=\pi a^2{I}_0$, the mutual inductance defined using this effective electric current isIntegrating an ideal magnetic dipole field through the entire disk would incorrectly use its singular expression inside the actual sphere; the rim calculation avoids that error.

For the prescribed time dependence, the quasistatic Faraday's law of induction result and the purely resistive loop approximation givePositive loop electric current is defined to produce a positive $z$ magnetic dipole moment, so the minus sign is Lenz's law. This assumes negligible retardation and $\nu L_b\ll R$, where $L_b$ is the loop's self-inductance. If that is not negligible, the governing equation is instead $L_b{\dot{I}}_b+R{I}_b=-{\dot{\Phi }}_b$; $R$ alone is insufficient for the exact dynamic response.

For static fields, the Maxwell equations give $\nabla \times B={\mu }_{0}J$ and $\nabla \cdot B=0$. Write $B=\nabla \times A$. A gauge transformation $A\mapsto A+\nabla \chi$ leaves $B$ unchanged; choose $\Delta \chi =-\nabla \cdot A$ to impose the Coulomb gauge. ThenFor a localized conserved current with the potential chosen to vanish at infinity, the supplied fundamental solution gives the magnetic vector potentialIts divergence vanishes by integration by parts and $\nabla \cdot J=0$. For a filamentary loop the current integral becomes $I{\oint }_{L}dr$, soFor $R=|x|$ much larger than the fixed loop size, expand $|x-r{|}^{-1}=R^{-1}+(x\cdot r)R^{-3}+O(R^{-3})$. The leading constant term integrates to zero because $\oint dr=0$. ThereforeThe vector triple-product identity gives $x\times (r\times dr)=r(x\cdot dr)-(x\cdot r)dr$. Integrating the derivative of $(x\cdot r)r$ around the closed loop shows $\oint r(x\cdot dr)=-\oint (x\cdot r)dr$. ConsequentlyEquivalently, $A={\mu }_{0}m\times x/(4\pi R^3)+O(R^{-3})$, where $m=(I/2)\oint r\times dr$ is the magnetic dipole moment. The leading displayed potential is of order $R^{-2}$; the remainder is smaller by a factor $R^{-1}$.

A steady localized electric current density satisfies $\nabla \cdot \mathbf{j}=0$, from charge conservation or the divergence of Ampère's equation. Choose an enclosing region whose boundary lies outside its support; extending the integration region does not change the source integrals. The divergence theorem applied componentwise givesTheir boundary fluxes vanish, proving the localized steady-current moment identitiesFor discontinuous but localized currents the same argument uses the distributional steady conservation law and an outer enclosing surface.

The magnetic vector potential is a vector field with $\mathbf{B}=\nabla \times \mathbf{A}$, consistent with $\nabla \cdot \mathbf{B}=0$. It is defined up to $\mathbf{A}\mapsto \mathbf{A}+\nabla \chi$. The steady Maxwell equation $\nabla \times \mathbf{B}={\mu }_0\mathbf{j}$ givesIn the Coulomb gauge $\nabla \cdot \mathbf{A}=0$, the equation is ${\nabla }^2\mathbf{A}=-{\mu }_0\mathbf{j}$, whose decaying Green-function solution is the one supplied in the source.

Let $r=|\mathbf{x}|$ and expand its kernel uniformly over the bounded source region:The first term integrates to zero. Put $T_{ki}=\int x_k^{\prime }{j}_i({\mathbf{x}}^{\prime })d{V}^{\prime }$, an antisymmetric tensor by the earlier identity. Define the magnetic dipole moment $m_{\ell }=\frac{1}{2}{ϵ}_{\ell ki}{T}_{ki}$, so $T_{ki}={ϵ}_{ki\ell }{m}_{\ell }$. ConsequentlyorThe displayed dipole term has size $r^{-2}$. If $\mathbf{m}=0$, it vanishes and a higher multipole determines the leading magnetic field instead.

For $r>0$, $\nabla \cdot (\mathbf{x}/r^3)=0$ andTherefore the far magnetic field is

Use SI units, $x^a=(ct,\mathbf{x})$ and metric signature $(+,-,-,-)$. The four-current is $j^a=(c\rho ,\mathbf{j})$. Charge density and current transform together under Lorentz transformations, and the continuity equation becomesWith the four-potential $A^a=(\phi /c,\mathbf{A})$, the Lorenz gauge ${\partial }_a{A}^a=0$ converts Maxwell equations to $\square A^a={\mu }_0{j}^a$, where $\square =c^{-2}{\partial }_t^2-{\nabla }^2$. Choosing no incoming homogeneous radiation, the retarded electromagnetic potential isCurrent conservation ensures this solution obeys the Lorenz gauge; a homogeneous solution may be added when additional field boundary data are specified.

For a thin loop, let $R=|\mathbf{x}|$, $\hat{\mathbf{R}}=\mathbf{x}/R$ and $t_R=t-R/c$. ThenIn the small-loop magnetic-dipole approximation, expand $|\mathbf{x}-{\mathbf{x}}^{\prime }|=R-\hat{\mathbf{R}}\cdot {\mathbf{x}}^{\prime }+\cdots$. The constant term integrates to zero because $\oint d{\mathbf{\ell }}^{\prime }=0$. The first nonzero term isFor the oriented circular loop the last integral is $\pi r^2\mathbf{n}\times \hat{\mathbf{R}}$. Thus with magnetic dipole moment $\mathbf{m}(t)=\pi r^{2}I(t)\mathbf{n}$,The $R^{-2}$ term is the near dipole potential and the $R^{-1}$ term its radiation potential. This is the leading small-$r$ result at fixed $R,\omega$; besides $r/R\ll 1$, the expansion of the retarded current requires $\omega r/c\ll 1$.

If only $r/R\ll 1$ is imposed and the loop need not be small compared with a wavelength, keep the retarded phase around the loop. For fixed $r,\omega$ and $R\to \infty$, the leading far-field expression is insteadwhere $\cos \theta =\mathbf{n}\cdot \hat{\mathbf{R}}$ and $J_1$ is the Bessel function of the first kind. Integrating the sinusoidal phase on a circle gives this Bessel factor; $J_1(z)\sim z/2$ recovers the radiation part of the boxed dipole result.

The microcanonical ensemble assigns equal probability to accessible quantum states in a narrow energy shell of an isolated system with fixed energy, particle number and external parameters. To derive the canonical ensemble, put a small system in weak thermal contact with a much larger isolated bath. If the small-system state has energy $E_i$, its probability is proportional to the number ${\Omega }_b(E_{\mathrm{total}}-E_i)$ of bath states. Expanding the bath entropy gives $S_b(E_{\mathrm{total}}-E_i)=S_b(E_{\mathrm{total}})-E_i/T+\cdots$, since $\partial S_b/\partial E=1/T$. Negligible interaction energy and an effectively constant bath temperature therefore give $p_i=e^{-E_i/(k_{B}T)}/Z$. The microcanonical ensemble suits an isolated fixed-energy system; the canonical ensemble suits a system exchanging energy with a thermostat at fixed temperature, with particle number still fixed.

For one atom, summing the finite geometric series over $m=-J,\dots ,J$ gives the partition functionAt $x=0$ the ratio is understood by continuity as $2J+1$. Logarithmic differentiation gives the mean magnetic dipole momentThis is odd in $x$, has small-$x$ expansion ${\mu }_{B}J(J+1)x/3+O(x^3)$, and saturates at $\pm J{\mu }_B$. The requested common graph is plotted in units of ${\mu }_B$: its ordinate is $⟨{\mu }_z⟩/(J{\mu }_B)$, so multiplication by ${\mu }_B$ gives $⟨{\mu }_z⟩/J$. The normalized slopes at zero are $(J+1)/3$ and the limits are $\pm 1$.

The magnetic susceptibility isEquivalently it is $N{\mu }_B^2\mathrm{Var}(m)/(k_{B}T)\ge 0$. For $k_{B}T\gg {\mu }_B|B|$, the apparent singular terms cancel, yielding Curie lawFor fixed nonzero $B$, $T\to 0$ and integer $J\ge 1$, the first excited magnetic level is separated from the aligned ground state by ${\mu }_B|B|$. Its exponentially small occupation givesThe moments are saturated, so changing the magnetic field hardly changes their alignment. This is a fixed-nonzero-field low-temperature limit. At exactly $B=0$, differentiating first gives the Curie expression for every $T>0$, which diverges as $T\to 0$; the two limits do not commute in this noninteracting model. For $J=0$ both moment and susceptibility vanish identically.

The magnetic torque transfers angular momentum from the faster rotator to the slower one. When ${\Omega }_m>{\Omega }_K(r_m)$, the neutron star supplies angular momentum to the inner accretion disk and possible propeller flow, so it tends to spin down. When the inner accretion disk rotates faster than the star, the magnetic connection and accreted angular momentum tend to spin it up.

For slowly varying mass accretion rate and magnetic dipole moment, these tendencies favour a magnetic spin equilibrium with the corotation radius near the magnetospheric truncation radius:The precise balance depends on accreted and ejected angular momentum and on the extent of magnetic coupling across the accretion disk; corotation is a scaling estimate, not an exact zero-torque theorem.

Align the polar axis with the magnetic dipole moment. The magnetic dipole field hasThe magnetic-field-line equation gives $dr/(rd\theta )=2\cot \theta$, hence the dipole magnetic-field line is $r=C{\sin }^2\theta$. Normalize the loaded bundle by its small surface angular radius ${\theta }_{\star }$ at $r=R_{\star }$. Its boundary obeysThe dipolar flux-tube area for one polar cap is consequentlyEquivalently, $BA$ is constant by conservation of magnetic flux and $B\propto r^{-3}$ near the axis. Two equal loaded caps double the total area. The $r^3$ scaling is independent of the normalization; if the loading angle is specified at another radius $r_L$, use $A(r)=\pi {\theta }_L^2{r}^3/r_L$ instead.

Let $R=|\mathbf{x}|$, $\hat{\mathbf{R}}=\mathbf{x}/R$, and $t_R=t-R/c$. To leading nonzero multipole expansion order, the loop is a time-dependent magnetic dipole withExpanding the retarded electromagnetic potential across the small loop givesThe $R^{-2}$ term is the near magnetic-dipole field and the $R^{-1}$ term is the magnetic dipole radiation field. Substituting the Fourier series for the current givesThe $n=0$ summand vanishes for the stated sine series.

Differentiate under the integral and use ${\nabla }_x|x-x^{\prime }{|}^{-1}=-{\nabla }_{x^{\prime }}|x-x^{\prime }{|}^{-1}$:The omitted boundary term vanishes if the current is localized sufficiently rapidly. A steady current obeys charge conservation ${\nabla }^{\prime }\cdot J=0$, hence

For $r=|x|$ much larger than the source size,Localization and ${\nabla }^{\prime }\cdot J=0$ imply $\int J{d}^3{x}^{\prime }=0$. They also implyby integrating ${\partial }_k^{\prime }(x_i^{\prime }{x}_j^{\prime }{J}_k)$. Thus the first nonzero moment is antisymmetric and can be written using the magnetic dipole momentConsequently the Coulomb-gauge vector potential of a localized steady current has far fieldThe dimensions are

Several historical models represented the same broad magnetic observations in different ways:

James Clerk Maxwell's later treatment explicitly contrasts the induction model associated with Siméon Denis Poisson with Weber's permanently magnetic molecules and discusses the predicted saturation limit.^[ref] Its 1873 date is retained; the chapter is used to explain the earlier hypotheses, not attributed to an 1865 publication.

Experiments constrain these models. Cutting a magnet produces smaller magnets with both poles, rather than simply separating free north and south charges. A current loop and an ordinary magnet also have closely related external dipole fields. Such facts explain the utility of molecular dipoles and currents, but do not uniquely identify the microscopic constituents. Modern magnetic domains, electron spin and quantum magnetic ordering refine the account; classical loops and freely turning isolated molecular magnets are incomplete microscopic pictures of a real ferromagnet. An effective pole calculation must likewise be distinguished from claiming an observed isolated magnetic monopole.

For Ørsted's experiment, place a long straight conducting wire close to and parallel with a freely moving horizontal compass needle. With no electric current, the needle follows the horizontal terrestrial magnetic field. Closing the circuit deflects it; reversing the electric current reverses the deflection, and moving the wire from above to below the needle also reverses the effect. For a long straight wire, the modern idealization iswith the circular direction given by the right-hand rule. These controls distinguish a magnetic action of the current from a fixed magnetized piece of wire or a mechanical disturbance.

Nobili's astatic galvanometer increases sensitivity in two ways. A coil with many turns multiplies the magnetic action of a given electric current. Two nearly equal magnetized needles are rigidly connected with opposite magnetic dipole moments and suspended by a fine thread. Their torques in the approximately uniform terrestrial field nearly cancel. In a common arrangement one needle lies inside the coil and the other above it, outside its strongest field, so the coil retains a substantial torque while the terrestrial restoring torque is reduced. A weak residual magnetic or suspension torque still provides a reproducible equilibrium; “astatic” means nearly cancelled terrestrial torque, not magnetic shielding. A museum instrument dated 1840 has many turns of fine wire and is identified with measurements of muscle currents.^[ref]

For weak physiological sources, use a long fine-wire winding with many turns, rather than the short thick winding appropriate to a low-resistance thermoelectric source. The physiological source commonly has a large effective internal resistance; a moderate added coil resistance is then outweighed by the gain in turns. This is a source-and-detector matching argument, not a claim that arbitrarily much wire always improves sensitivity.

For example, at fixed mean turn length and fixed copper winding volume, $R_{\mathrm{coil}}=a{N}^2$. For source electromotive force $\mathcal{E}$ and resistance $R_s$, the small-angle driving torque is proportional toDifferentiation gives its maximum at $\boxed{R_{\mathrm{coil}}=R_s}$. This simple optimal turn count for fixed winding volume model explains why high-source-resistance work favors more turns, while also exposing the limit. Actual instruments additionally trade response time, damping, geometry and source loading against their steady deflection.

Only the magnetic-response physics is addressed here; biological and clinical application is user-deferred.

A single-domain particle has an approximately coherent magnetic dipole moment, but single-domain does not by itself imply rapid thermal reversal. If moments equilibrate within the measurement time, a dilute, noninteracting ensemble is superparamagnetic: its quasistatic magnetization is reversible, odd in field and approaches saturation with negligible remanence or magnetic coercivity.

For freely orientable classical moments,At small field, $\mathcal{L}(z)\simeq z/3$; at large field moments align. This Langevin form is an ideal limit, not a universal exact curve for anisotropic fixed-axis particles.

The missing timescale/magnetic anisotropy information must be stated: blocked single-domain particles can already show hysteresis. Also, a reversible quasistatic curve does not preclude a finite-frequency loss loop when relaxation lags an alternating field. The original curves illustrate an equilibrated dilute limit and possible blocked/interacting alternatives, not measured treatment performance.

Spin-orbit coupling is a relativistic interaction between a particle's spin and its orbital angular momentum or, more generally, its motion through an electric field. A moving electron experiences a transformed magnetic field, which couples to its magnetic dipole moment. For an electron bound in a central electrostatic potential energy $U(r)$, the leading interaction, including the Thomas precession correction, isHere $\mathbf{L}$ is orbital angular momentum and $\mathbf{S}$ is spin. The factor of one-half relative to the simplest rest-field argument comes from Thomas precession of the accelerated electron frame.^[ref]

For example, in the hydrogen atom, the $2p$ states with total angular momentum $j=1/2$ and $j=3/2$ have different $\mathbf{L}\cdot \mathbf{S}$ eigenvalues. Their splitting is part of the atomic fine structure. The interaction redistributes and splits spin-dependent energies; it is not the same physical effect as the spin-independent convective magnetic force between charged filaments.

The information in the original question is insufficient to determine a unique spin-induced change of charge density. It specifies ideal dielectric filaments with a prescribed excess linear charge density, but not the charge carrier species, their spin state, charge carrier mobility, binding potential or relaxation mechanism. A classical line charge has no spin variable at all. The following conditional conclusions are supported by the fields already calculated.

For a carrier moving with its filament, the local magnetic field in its instantaneous rest frame is, for the transverse fields here,where $\mathbf{u}$ is that carrier's laboratory velocity. Exclude its filament's singular self-field and first retain only the prescribed straight convective motion.

In Case 1 there is a common rest frame in which both filaments and the excess charges are stationary and the interfilament magnetic field vanishes. Equivalently, in the laboratory ${\mathbf{B}}_A=\mathbf{v}\times {\mathbf{E}}_A/c^2$, so B's co-moving charges have ${\mathbf{B}}_{A,\mathrm{rest}}=0$. Thus uniform common translation alone creates no extra mutual rest-field spin splitting. This does not exclude intrinsic spin-orbit coupling of electrons moving within their atomic binding potentials; that material physics was never specified.

In Case 2 there is relative motion and no frame in which both source distributions are stationary. Put $E_{\ast }=\eta /(2\pi {ϵ}_{0}d)$ as in part (c). At the respective filament axes, the external fields experienced in each carrier distribution's own rest frame areFor the first expression B is at rest in $O$; for the second A is at rest in $O^{\prime }$ and the field comes from B moving at $-v\hat{\mathbf{x}}$. For a carrier with magnetic dipole moment ${\mathbf{\mu }}_s$, the interaction $H_s=-{\mathbf{\mu }}_s\cdot {\mathbf{B}}_{\mathrm{rest}}$ distinguishes opposite transverse spin orientations. A field gradient could exert a spin-dependent force $\nabla ({\mathbf{\mu }}_s\cdot {\mathbf{B}}_{\mathrm{rest}})$ on mobile carriers; spin relaxation could instead produce a spin polarization. These are possible responses, not consequences of charge density alone. In B's frame the displayed interaction is directly a magnetic-moment coupling to the moving source's field; treating the moving carrier's electrostatic orbital term in another frame also requires consistent binding and acceleration dynamics, including any applicable Thomas precession.

For example, if $\eta >0$ and mobile electron spins can relax, the lower magnetic-energy orientation has the electron's magnetic moment toward $+\hat{\mathbf{y}}$ and its spin toward $-\hat{\mathbf{y}}$. Oppositely oriented moments would experience opposite gradient forces. The sign changes if the source charge changes sign. None of these statements fixes a net charge redistribution without a confinement and response model, and the stipulated zero transverse size supplies no resolved “sides” on which charge could accumulate.

Therefore Case 1 has no new mutual spin effect from common bulk motion alone; Case 2 permits motion-induced spin-dependent interactions, but neither case determines an additional charge-density profile from the supplied data. In particular, the factor $\gamma$ relating proper and laboratory linear charge densities in part (b) follows from length contraction and is not an effect of spin-orbit coupling. If the charges remain pinned in an ideal dielectric, their imposed density remains the prescribed density in either case.

The separate proton and neutron shell capacities are $2j+1=2,4,2,6,\dots$. Filled shells and paired particles have zero net spin and magnetic dipole moment. The single-particle nuclear magnetic moment is $\mu =g_{j}j{\mu }_N$, with orbital factors $g_{\ell }^p=1$, $g_{\ell }^n=0$ and $s=1/2$.

For nitrogen, the eight neutrons form a closed shell. The seventh proton occupies $1p_{1/2}$, so $\ell =1$, $j=1/2$ and the Landé g-factor givesHence the nuclear shell model predictsIts single-particle spin and parity is $1/2^{-}$. The negative moment results from the proton spin being predominantly opposed to the total angular momentum in this $j=\ell -1/2$ level.

For oxygen, eight protons and the first eight neutrons are paired. The ninth neutron occupies $1d_{5/2}$, with $\ell =2$, $j=5/2=\ell +1/2$. Since its orbital factor is zero, $g_j=g_n/(2\ell +1)=-3.826/5$. ThereforeThese are ideal single-particle predictions in units of the nuclear magneton, not fits incorporating nuclear configuration mixing.

The magnetic dipole moment is $\mathbf{\mu }=IA\hat{\mathbf{n}}$ and its magnetic torque is ${\mathbf{\tau }}_{\mathrm{mag}}=\mathbf{\mu }\times \mathbf{B}$. Its component along the rotation axis opposes the imposed rotation. For positive $\omega$, the external driving torque needed to keep the angular velocity constant isAveraging over a complete rotation, with $⟨{\sin }^2\omega t⟩=1/2$, givesAs an independent sign and dimension check, $\omega ⟨{\tau }_{\mathrm{drive}}⟩=R⟨I^2⟩$. Thus the mechanical input power equals the mean Joule heating; a passive loop cannot supply a sustained driving torque. This average assumes the same negligible-self-inductance approximation as the current expression.

The elemental torque about the origin is $d\mathbf{\tau }=\mathbf{r}\times d\mathbf{F}$. Substitution of part (a) givesOnly its $x$ component drives the allowed rotation about the $x$ axis. Using ${\int }_0^{2\pi }{\sin }^2\theta d\theta =\pi$ and ${\int }_0^{2\pi }\sin \theta \cos \theta d\theta =0$, the projected torque isThe magnetic dipole moment is $\mathbf{m}=I\pi a^2\hat{\mathbf{z}}$, so the full electromagnetic torque isIts $x$ component agrees exactly. The $y$ component is a real electromagnetic torque, but the bearing constraining motion to the $x$ axis supplies a cancelling reaction torque. For an allowed infinitesimal rotation $d\mathbf{\alpha }=\hat{\mathbf{x}}d\alpha$, only $\mathbf{\tau }\cdot d\mathbf{\alpha }={\tau }_{x}d\alpha$ does mechanical work.