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Assume the electrical conductor is nonmagnetic, with magnetic permeability ${\mu }_0$, and neglect end effects. Rotational symmetry and translational symmetry make the magnetic flux density azimuthal and constant along a coaxial circle. The magnetostatic Ampère law therefore givesUniform current density is $j=I/(\pi a^2)$, so $I_{\mathrm{enc}}(r)=j\pi r^2=I{r}^2/a^2$. ThusThe direction follows the right-hand rule around the electric current. For a uniform linear magnetic material, replace ${\mu }_0$ by its magnetic permeability.

Stokes theorem converts circulation around an oriented boundary into curl flux through a spanning surface. For a sufficiently smooth vector field on an oriented piecewise smooth surface $S$,The boundary orientation follows the right-hand rule relative to $\mathbf{n}$. The field must be smooth on the surface, not merely on its boundary; a singularity crossing the surface cannot be ignored. For a fixed loop, applying the theorem to Faraday's differential law gives the integral flux law of induction.

The planar form is Green theorem. For a region $D$ with positively oriented boundary and sufficiently smooth $P,Q$,The outer boundary is counterclockwise and any holes have clockwise boundaries. For instance $P=-y/2,Q=x/2$ makes the area of $D$ equal to half the line integral of $xdy-ydx$; the unit circle gives area $\pi$.

A closely related use of “Green's theorems” in field theory is Green's identities, obtained by applying the divergence theorem to $\varphi \nabla \psi$:and, after subtracting the expression with $\varphi ,\psi$ exchanged,These relate interior sources to boundary data in electrostatics and other boundary-value problems. For example, a harmonic potential difference vanishing on a Dirichlet boundary condition has zero integral of its squared gradient by the first identity, establishing uniqueness. Both the planar circulation theorem and the field identities are included here to make the convention clear.

Choose the plate polarity so that the gap electric field points along $+z$ when $V>0$. Away from the plate edges, the quasistatic electromagnetic approximation gives $E_z=V_0\sin \omega t/d$. A circular Ampere-Maxwell law contour of radius $r<a$ then givesIt is displacement current, rather than a conduction current across the air, that supplies the enclosed current term. The direction follows the right-hand rule for increasing positive electric flux; it reverses every half cycle. In the ideal extension that sets the gap electric field to zero outside the plate footprint, $B_{\varphi }={\mu }_0{ϵ}_0\omega V_0{a}^2\cos \omega t/(2dr)$ for $r>a$. That extension is an edge approximation, not an exact exterior solution for finite plates. Neglect fringing field of a capacitor and retardation, requiring $d\ll a$ and $\omega a/c\ll 1$.

The Biot-Savart law gives the magnetic field of a steady thin-wire electric current in vacuum:The cross product fixes the circulation direction by the right-hand rule; fields from different electric current elements superpose. Integration gives $B={\mu }_{0}I/(2\pi r)$ outside a straight infinite wire and $B={\mu }_{0}I/(2a)$ at the centre of a circular loop of radius $a$. A distributed electric current replaces $Id{\mathbf{\ell }}^{\prime }$ by $\mathbf{J}d{V}^{\prime }$. This is a magnetostatic law: arbitrary time-dependent electric currents require retarded electromagnetic fields, not instantaneous substitution.

The differential form of Faraday's law of induction isBy Stokes theorem, for a stationary oriented circuit it becomes $\mathcal{E}=\oint \mathbf{E}\cdot d\mathbf{\ell }=-d{\Phi }_B/dt$, where ${\Phi }_B=\int \mathbf{B}\cdot d\mathbf{S}$ is magnetic flux and $\mathcal{E}$ is electromotive force. Circuit orientation and surface normal obey the right-hand rule. Lenz's law is the physical meaning of the minus sign: the induced current's magnetic field opposes the change in linked magnetic flux. For a moving circuit, the force per unit electric charge is $\mathbf{E}+\mathbf{v}\times \mathbf{B}$, and its line integral is the corresponding electromotive force. Motion through a static magnetic field can therefore induce a voltage even when ${\partial }_t\mathbf{B}=0$.

Let $\theta$ be the wire's angle from the downward vertical, $r=L\sin \theta$, and choose $\mathbf{\omega }$ and $\mathbf{B}$ upward, as in the figure. The motional term along the wire has radial component $B\omega s$ at cylindrical radius $s$. In steady electrical equilibrium, integrating from the grounded axis to the bob givesThe Lorentz force on this positive electric charge is radially outward. Its ratio to the required centripetal term is $QB/(m\omega )\lesssim C{B}^2{L}^2/(2m)$, so the stated heavy-bob condition makes its effect on the zero-field conical pendulum geometry negligible. With wire tension $T_w$, $T_w\cos \theta =mg$ and $T_w\sin \theta =m{\omega }^{2}r$, giving $\cos \theta =g/({\omega }^{2}L)$. Hence the magnetically charged conical pendulum hasA nonzero-radius orbit requires ${\omega }^{2}L>g$. Reversing the sense of rotation or the magnetic field reverses the sign of $Q_0$.

The response to the added electric field requires an electrical convention. In the usual mechanical approximation, treat the pre-existing electric charge as $Q_0$, include its upward electric force $Q_{0}E$, and neglect all magnetic-force feedback. ThenThis predicts an increased radius for $Q_{0}E>0$. It is the result when the newly applied field is used only as a probe force on the original electric charge.

There is a first-order qualification if the bob remains electrically connected to the grounded pivot and the uniform added field's potential is included consistently. Put $s=\cos \theta$ and set the external electric potential to zero at the pivot. The external potential at the bob is then $ELs$, so the charging relation and full force balance becomeThe second equation follows by eliminating the wire tension between vertical balance and radial balance, including the outward magnetic Lorentz force. Expanding both equations to first order in $E$, and to leading order in $C{B}^2{L}^2/m$, with $s_0=g/({\omega }^{2}L)$, givesThe extra factor arises from the added field changing $Q$ by $-CEL{s}_0$ and therefore changing the radial magnetic force. That term is linear in $E$ too; the heavy-bob assumption alone does not justify retaining $Q_{0}E$ while dropping it. The two expressions refer to the two electrical approximations just specified. The expansion requires a small change relative to the initial radius and is nonuniform as that radius tends to zero.

For an electrical measurement, route the grounded connection through a sensitive electrometer operated near zero input voltage, and integrate the charging electric current when $B$ or $\omega$ is changed. Starting with the field off, the transferred electric charge is $Q_0$ in this ideal capacitance model. Reversing $B$ separates the induced signal from offsets. Alternatively, isolate the charged bob briefly and measure its electric potential with a high-input-impedance electrometer, accounting for the instrument's extra capacitance and for the fact that isolation changes the electrical boundary condition.

At distance $r\gg w$, the strip is indistinguishable to leading order from a straight line current. Ampère's law on a large circular contour gives $B\simeq {\mu }_{0}I/(2\pi r)$, directed tangentially around the strip's longitudinal axis by the right-hand rule. On the perpendicular centreline this also follows by expanding ${\tan }^{-1}(w/(2z))\simeq w/(2z)$.

Field law and centre value. The Biot-Savart law for a steady thin wire isHere $d{\mathbf{\ell }}^{\prime }$ follows the conventional electric current and the displacement points from source to observation point. For a circular loop of radius $a$, every element at the centre has distance $a$ and contributes in the same axial direction. Choose $+z$ by the right-hand rule. Then

On-axis integration. At axial distance $z$, every wire element has distance $\sqrt{a^2+z^2}$. Azimuthal symmetry cancels transverse components. Parameterizing ${\mathbf{r}}^{\prime }=(a\cos \phi ,a\sin \phi ,0)$ gives axial cross-product element $a^{2}d\phi$, so the on-axis magnetic field of a circular current loop isIt points the same way on the two halves of the axis; its strength is even in $z$.

Directions near the axial point. At a point on the positive axis the field is axial, but nearby lines spread outward because $B_0$ decreases as $z$ increases. On the negative-axis side they converge toward the axis while still progressing in the positive axial direction. There is no azimuthal field. The near-axis sketch shows this local spreading; the complete magnetic field lines return around the outside of the loop and form closed curves.

Radial component from zero divergence. Smooth axial symmetry gives $B_z(\rho ,z)=B_0(z)+O({\rho }^2)$ and $B_{\rho }=O(\rho )$. By Gauss's law for magnetism,Integrate radially from zero, excluding a singular $1/\rho$ term, to obtain the near-axis radial magnetic field, $B_{\rho }=-\rho B_0^{\prime }(z)/2+O({\rho }^3)$. Differentiating the axial field therefore givesFor $z>0$ it is outward; for $z<0$ it is inward. This supplies the direction reasoning used in the sketch.

Force on the second loop. Put the second loop at $z=d$ and radius $\rho =a$. For $d\gg a$, the radial field due to the first loop is $B_{\rho }\simeq 3{\mu }_{0}I{a}^3/(4d^4)$. The element $d\mathbf{\ell }=ad\phi \hat{\mathbf{\phi }}$ experiences Lorentz forceThe radial contributions cancel as vectors around the loop. The axial contributions add, giving the force between distant coaxial current loops,The sign follows $\hat{\mathbf{\phi }}\times \hat{\mathbf{\rho }}=-\hat{\mathbf{z}}$. Both the small radial coordinate relative to $d$ and the dipole approximation are justified by $a/d\ll 1$.

Force direction. The second loop is pulled back toward the first: the force is attractive for the same current sense. Reversing one current reverses the force. The much larger axial field chiefly produces radial element forces; retaining the smaller radial field is essential for the net axial attraction.

The field law. In the Biot-Savart law, $d\mathbf{B}$ is the contribution to the magnetic flux density at the observation point from a directed wire element $d\mathbf{\ell }$ carrying steady conventional electric current $I$. The vector $\mathbf{R}$ points from that element to the observation point, $R=|\mathbf{R}|$, and $\hat{\mathbf{R}}=\mathbf{R}/R$. The constant ${\mu }_0$ is the vacuum permeability; the cross product sets the direction by the right-hand rule. Integrating the elements gives the total field, in teslas.

The moment of the shown loop. The arrows in the original PDF run counterclockwise viewed from $+z$. Thus the magnetic dipole moment isIts units are $\mathrm{A}{\mathrm{m}}^2$. Reversing the current reverses both this moment and the field.

Induction and its sign. The Maxwell-Faraday equation is $\nabla \times \mathbf{E}=-{\partial }_t\mathbf{B}$. For a fixed loop, Stokes theorem givesThis is Faraday's law of induction. The orientation of the surface normal and positive circulation are linked by the right-hand rule. The minus sign is Lenz's law: the induced current's linked flux opposes the imposed flux change, rather than reinforcing it.

The inductive current. Let the loop normal make angle $\theta$ with the applied field, with positive current defined by that normal. The external magnetic flux is ${\Phi }_{\mathrm{ext}}=A{B}_0\sin \omega t\cos \theta$. With negligible electrical resistance, the circuit equation is $L\dot{I}=-{\dot{\Phi }}_{\mathrm{ext}}$, so total linked flux $LI+{\Phi }_{\mathrm{ext}}$ is constant. For zero initial trapped flux,The printed result needs this initial-flux assumption; in general an additional constant current is allowed. The units are amperes because a weber divided by a henry is an ampere.

Current and forces at one instant. Take $B>0$ along $+x$ and $0<\theta <\pi /2$, so $I<0$ relative to the chosen normal. The loop magnetic dipole moment $\mathbf{\mu }=IA\mathbf{n}$ therefore has a component opposite the applied field. Each wire element experiences Lorentz force $d\mathbf{F}=Id\mathbf{\ell }\times \mathbf{B}$. Opposite sides have equal and opposite forces, giving zero net force in the uniform field but a nonzero couple.

The horizontal top and bottom wire segments also experience equal and opposite vertical forces. They cancel without adding torque about the vertical rotation axis; the displayed force pair supplies that torque.

Average torque. Since $\mathbf{\tau }=\mathbf{\mu }\times \mathbf{B}$, its generalized component for increasing $\theta$ is ${\tau }_{\theta }=-IAB\sin \theta$. Substitution givesand, using $⟨{\sin }^2\omega t⟩=1/2$,The orientation torque on a flux-conserving loop drives its normal away from the field for the stated initial angle. Its magnitude has units joules, or newton metres.

Free motion with large inertia. Write $J$ for the mechanical moment of inertia. If its orientation changes little during each field cycle, the slow equation is $J\ddot{\theta }=⟨{\tau }_{\theta }⟩$, with effective potentialThe stable orientation is $\theta =\pi /2$, where the loop plane contains the applied field. Released from rest between $0$ and $\pi /2$, the loop accelerates toward that orientation, overshoots and librates around it. In the leading averaged, undamped model its turning angles are ${\theta }_i$ and $\pi -{\theta }_i$. It does not settle without a dissipative mechanism. For small librations the squared angular frequency is $A^2{B}_0^2/(2LJ)$.

Negligible inductance and finite resistance. Now the current is proportional to the induced electromotive force, not to the flux. Including slow rotation,The rotational damping of a resistive loop in an oscillating field follows fromAt fixed orientation, the first term alternates at $2\omega$ and has zero mean. The loop can therefore rock slightly, but there is no leading mean aligning torque like that in the inductive case. Slow rotation giveswhich opposes motion and produces Joule heating. Initial-phase transients can initiate motion, and the instantaneous torque is not zero. This description retains the large-inertia, slow-motion approximation; it does not assert an exact stationary solution at every starting angle.

For a steady filamentary electric current $I$ in vacuum, the Biot-Savart law isThe line element points with conventional electric current, and the cross product fixes the magnetic flux density direction by the right-hand rule.

As an application, an infinitely long straight wire has circular symmetry about its axis. At perpendicular distance $s$,Its direction follows the right-hand rule. The following part applies the same law to a circular current loop.

Apply Ampère's law on a concentric circle. Uniform current density is $J=I/(\pi a^2)$, so the enclosed current is $I{r}^2/a^2$ inside and $I$ outside. With permeability ${\mu }_0$,The field circulates azimuthally by the right-hand rule. Its magnitude rises linearly from zero, is continuous and maximal at the surface, then falls as $1/r$. End effects are neglected.

An electric dipole consists of opposite charges separated by a small displacement, with electric dipole moment $\mathbf{p}=q\mathbf{d}$ directed from negative to positive. At distances large compared with its size,The net charge term vanishes; the field decays faster than a monopole field. In an imposed slowly varying electric field,for a permanent fixed dipole. A uniform field produces a torque but no net force. For a linear induced dipole $\mathbf{p}=\alpha \mathbf{E}$, the interaction energy is instead $-\alpha E^2/2$, because building the dipole also costs energy.

A small planar current loop has magnetic dipole moment $\mathbf{\mu }=IA\hat{\mathbf{n}}$, with the normal fixed by the right-hand rule. Outside the source,Magnetic field lines form closed loops; a current-loop dipole does not require magnetic monopoles. Dipole moments describe the leading directional field of a neutral localized source, and their torques explain alignment in external fields.

Electromagnetic induction is the production of an electromotive force by changing magnetic flux or by moving a conductor through a magnetic field. With the loop and surface orientations related by the right-hand rule, Faraday's law of induction isHere $\mathbf{v}$ is the local wire velocity. For a stationary loop, the induced electric field can have a nonzero closed-path integral; its differential form is $\nabla \times \mathbf{E}=-\partial \mathbf{B}/\partial t$. For a straight rod moving perpendicularly to itself and to a uniform magnetic field, the motional electromotive force has magnitude $B\ell v$.

The minus sign expresses Lenz's law: the induced electric current produces a magnetic field that opposes the change in linked magnetic flux. Pulling a conducting loop through a magnetic field therefore requires work against an electromagnetic resisting force; that work supplies the resulting heat or stored energy.

For an inductor with constant self-inductance, $\mathcal{E}=-LdI/dt$ and its stored magnetic energy is $L{I}^2/2$. A changing electric current in one coil can induce an electromotive force in another through mutual inductance, the operating principle of a electrical transformer. These examples connect the flux law, the direction of the response and conservation of energy.

Directions. Choose $+x$ along the horizontal spindle from the pivot to the disk, and $+z$ vertically upward. With lever arm $\mathbf{r}=l\hat{\mathbf{x}}$ and weight ${\mathbf{F}}_g=-mg\hat{\mathbf{z}}$, the torque about the pivot isClockwise precession viewed from above has $\mathbf{\Omega }=-\Omega \hat{\mathbf{z}}$. The relation $\mathbf{\tau }=\mathbf{\Omega }\times \mathbf{L}$ therefore requires the spin angular momentum to point along $-x$, from the disk toward the pivot. The disk spins clockwise when viewed from its outer face looking along the spindle toward the pivot. Looking from the pivot reverses that apparent rotation. The right-hand rule fixes these signs unambiguously.

The drawing shows an instantaneous top view; the disk's angular momentum and the horizontal spindle turn clockwise together. The pivot force has zero torque about the pivot.

Rate of precession. For steady horizontal motion the spin contribution has magnitude $L=I\omega$ and turns through $d\varphi =\Omega dt$ in time $dt$. Its change is perpendicular to itself and has magnitude $dL=Ld\varphi$. HenceThis is the usual slow steady gyroscope precession relation. Any constant vertical angular momentum associated with steady precession has zero time derivative; the horizontal spin contribution supplies the changing component here. Transient nutation is excluded by the prescribed horizontal steady motion.

Clock design. Convert both rotation rates to angular frequencies:For the moment of inertia of a uniform disc, $I=m{R}^2/2$, so the mass cancels:

Spin friction. At fixed geometry, reducing $\omega$ increases $\Omega \propto 1/\omega$. The precession period, $T_p=2\pi /\Omega$, therefore decreases. This trend holds while slow steady precession remains a suitable description; it cannot be extrapolated to a stopped disk.

Counterweight. To oppose the disk's gravitational torque, place the counterweight on the other side of the pivot, at $-l_c\hat{\mathbf{x}}$. The net torque isA horizontal axle can remain fixed when this vanishes, givingFor the fixed spin direction already found, counterbalanced gyroscope precession has signed rateMoving the counterweight outward from the pivot first slows the clockwise motion, eliminates its driving torque at balance, and then reverses the steady motion to counterclockwise. A counterweight on the disk's side would add to the gravitational torque. These statements neglect axial spin moment of inertia of the counterweight and assume the stated horizontal motion, without a separately imposed precession at balance.

Faraday's law of induction relates an induced electromotive force to the rate of change of magnetic flux. Choose a positive orientation around a loop and its associated surface normal by the right-hand rule. For a stationary conducting loop,Unlike an electrostatic field, the induced electric field generally has nonzero curl; the differential form is $\nabla \times \mathbf{E}=-\partial \mathbf{B}/\partial t$. For a moving circuit, magnetic flux transport for a moving circuit includes the magnetic Lorentz force on the charges:where $\mathbf{u}$ is the local wire velocity. Thus changing the field or moving a conductor can produce induction.

The minus sign is Lenz's law: an induced electric current, when a closed conducting path permits it, produces a magnetic field opposing the change of linked flux. It need not oppose the original field itself. For example, increasing flux through a loop produces a current tending to reduce that increase; decreasing the same flux reverses the current. The opposition is consistent with conservation of energy: external mechanical or electrical work is needed to maintain a change against its inductive reaction.

For a coil of $N$ turns, define the magnetic flux linkage $\Lambda ={\sum }_{j=1}^N{\Phi }_j$, reducing to $N{\Phi }_B$ when every turn links the same flux. For fixed geometry and a linear magnetic response, self-inductance $L$ is defined by $\Lambda =LI$, and the coil's induced electromotive force isThe corresponding terminal voltage drop, with the passive sign convention, is $V_L=LdI/dt$. It opposes changes in electric current; it does not require the current itself to vanish.

Building the current from zero requires power $P=I{V}_L=LIdI/dt$. Integrating gives the energy stored by an ideal inductor:This energy is stored in the magnetic field and can be returned when the current falls. In a series RL circuit driven by a constant voltage $V_0$, Kirchhoff's voltage law gives $L\dot{I}+RI=V_0$. Starting with zero current, $I=(V_0/R)(1-e^{-tR/L})$, with time constant $L/R$. This makes the resistance to rapid current changes directly observable. Two coils also have mutual inductance: changing primary current induces secondary electromotive force $-Md{I}_1/dt$ with a polarity determined by the chosen winding orientations, the principle behind a transformer.

Choose a positive unit normal $\hat{\mathbf{n}}$ to the loop and choose positive circulation by the right-hand rule. Write the applied magnetic field as $\mathbf{B}(t)=B_0\cos (\omega t)\hat{\mathbf{b}}$, where $\hat{\mathbf{b}}\cdot \hat{\mathbf{n}}=\cos \theta$. Because the field is uniform, the external magnetic flux isFaraday's law of induction gives the induced electromotive force, and Ohm's law gives the current through the loop's electrical resistance:This is the induced current in a tilted resistive loop with the self-generated magnetic-flux contribution neglected as requested. The factor has units $\mathrm{Wb}{\mathrm{s}}^{-1}{\mathrm{\Omega }}^{-1}=\mathrm{A}$. A field parallel to the loop plane has $\cos \theta =0$ and induces no current in this uniform-field model.

For $\cos \theta >0$, the initial normal flux is positive. At $\omega t=\pi /2$, the external magnetic field is instantaneously zero but its normal component is decreasing, so $d{\Phi }_B/dt<0$ and $I>0$. The current is counterclockwise when viewed from the positive-normal side. Its own magnetic field points along $+\hat{\mathbf{n}}$ through the loop, preserving the initial normal flux in accordance with Lenz's law.

For $\cos \theta <0$, both circulation and the induced normal-field direction reverse. The current follows the rate of change of magnetic flux; the quarter-cycle current is nonzero even though the external field itself vanishes then.

For steady electric currents in vacuum, Ampère's circuital law isChoose an oriented closed contour and a spanning surface; the positive enclosed electric current crosses that surface according to the right-hand rule for the contour. First use the source symmetry to establish the direction of $\mathbf{B}$ and which positions have the same magnitude. Then choose the contour so its useful segments are parallel to $\mathbf{B}$ with constant magnitude, and the remaining segments are perpendicular to $\mathbf{B}$ or have vanishing field contribution. The line integral reduces to the unknown magnitude times a known length or sum of lengths.

For example, cylindrical symmetry around an infinite wire allows a circular contour, giving $B(2\pi r)={\mu }_{0}I$. A uniform current sheet instead calls for a rectangle straddling it. The law is always valid in this magnetostatic setting, but by itself does not determine the local magnetic field from an arbitrary enclosed current without these extra symmetry arguments. Here ${\mu }_0$ is the vacuum permeability.

Choose $+x$ along the electric current, $+z$ upward normal to the sheet, and $+y$ to complete a right-handed frame. In the interior region far from its edges, treat the sheet as infinite in its plane. Its surface current density isReflection and translational symmetry make the magnetic field parallel to the sheet, perpendicular to its current, and equal in magnitude with opposite direction above and below. Choose an Ampère's circuital law rectangle with length $s$ along $y$, straddling the full thickness. The two long sides each contribute $Bs$ with the same sign; on the vertical sides $\mathbf{B}\cdot d\mathbf{l}=0$ since $\mathbf{B}$ is horizontal. The enclosed electric current is $Ks=Is/w$. ThusThe right-hand rule fixes the directions:The current density inside the finite-thickness conductor is $\mathbf{J}=I/(wt)\hat{\mathbf{x}}$, so integrating through thickness $t$ gives the same $\mathbf{K}$. These are exterior fields in the broad-sheet approximation, not exact finite-width edge fields.

Translation and rotation. For particles constituting a rigid body, the total linear momentum isThus the total external force governs the motion of the centre of mass. About that centre of mass, the angular momentum and applied torque arewhere primes denote relative positions and velocities. The derivative of the vector is taken relative to nonrotating axes. The inertia tensor relates angular momentum to angular velocity, ${\mathbf{L}}_{\mathrm{CM}}=\mathsf{I}\mathbf{\omega }$. For rotation about a fixed unit direction $\mathbf{n}$, the moment of inertia is $I_{\mathbf{n}}=\int r_{\perp }^{2}dm=\mathbf{n}\cdot \mathsf{I}\mathbf{n}$; it measures the resistance to changing that rotation and gives rotational kinetic energy $I_{\mathbf{n}}{\omega }^2/2$. The simpler vector relation $\mathbf{L}=I\mathbf{\omega }$ applies when the rotation axis is a principal axis, not for every orientation of an arbitrary rigid body. These translational and rotational balances must be applied together; a nonzero torque does not replace the net-force equation.

Gyroscope precession. A rapidly spinning gyroscope has a large angular momentum approximately along its spin axis. The torque component parallel to this vector changes its magnitude, while the perpendicular component turns its direction. If the vector undergoes precession with angular velocity $\mathbf{\Omega }$, then over $dt$ its change is the infinitesimal rotation $d\mathbf{L}=(\mathbf{\Omega }dt)\times \mathbf{L}$. HenceThis describes direction-changing precession at constant $|\mathbf{L}|$; any spin-up contribution must be added separately. For perpendicular torque and angular momentum, the precessional component perpendicular to $\mathbf{L}$ is ${\mathbf{\Omega }}_{\perp }=\mathbf{L}\times \mathbf{G}/L^2$. The familiar tendency of a supported gyroscope to turn sideways is this vector balance, not the absence of gravitational force.

Wheel-lift threshold without the flywheel. Let $h=0.50\mathrm{m}$ be the centre of mass height, $a=1.0\mathrm{m}$ half the wheel separation, and $R_b=10\mathrm{m}$ the bend radius. The required inward force is $F=M{v}^2/R_b$, provided by static friction. Let $N_o,N_i$ be the summed outer and inner wheel normal forces. Vertical balance gives $N_o+N_i=Mg$. Moment balance about the centre of mass givesThe inner wheels lift when $N_i=0$. Thus the vehicle rollover threshold isThe stated mass cancels. This assumes the rigid geometry and sufficient static friction specified in the paper; the threshold is set by loss of contact, not by sliding.

Engine flywheel. The moment of inertia of a uniform disc givesIts spin angular speed, spin angular momentum, and turn angular speed at $40\mathrm{km}{\mathrm{h}}^{-1}$ areThe transverse flywheel axis is perpendicular to the vertical turn axis, soTo specify its direction, use right-handed axes $+x$ forward, $+y$ to the car's left, and $+z$ upward, and consider a left turn. Then $\mathbf{\Omega }=+\Omega \hat{\mathbf{z}}$. Choose the flywheel angular momentum to the car's right, ${\mathbf{L}}_f=-L_f\hat{\mathbf{y}}$. The car must applyto obtain the required precession. This gyroscopic roll torque in a turning vehicle opposes the outward roll tendency in a left turn. By the right-hand rule, the flywheel must turn counterclockwise viewed from the car's right side toward its centre, opposite to the rotation of wheels rolling forward. Reversing flywheel spin would aggravate the outward roll. Reversing the turn reverses both roll and gyroscopic torques, so the same flywheel spin remains helpful.

For comparison, the outward inertial roll moment at the stated speed is $Mh{v}^2/R_b=6173\mathrm{N}\mathrm{m}$, and the limiting gravitational restoring moment is $Mga=9810\mathrm{N}\mathrm{m}$. The gyroscopic reaction is about $30\%$ of the former and $19\%$ of the latter, so it is appreciable for this large flywheel. If this ideal model is extended to include the favorable gyroscopic reaction in the wheel-lift condition, it gives $Mh{v}^2/R_b-L_{f}v/R_b=Mga$, or about $56.8\mathrm{km}{\mathrm{h}}^{-1}$ rather than $50.4\mathrm{km}{\mathrm{h}}^{-1}$. This quantifies a useful but finite increase in the rollover threshold.

For an infinite straight wire in vacuum, Ampère's circuital law around a circle of radius $R$ gives $B(2\pi R)={\mu }_{0}I$. ThereforeHere ${\mu }_0$ is the vacuum permeability and the azimuthal direction follows the right-hand rule around the electric current. The magnetic field is measured in $\mathrm{T}$.