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For a steady axisymmetric compressible fluid flow, mass conservation permits this representation. The mass flow through an annulus between two stream surfaces is $2\pi \Delta \psi$. In a steady adiabatic wind without magnetic forces, the specific entropy, specific angular momentum and Bernoulli function are locally functions of $\psi$ on connected regular stream surfaces.

The inviscid magnetohydrodynamic momentum equation balances inertia, pressure, gravitational acceleration and Lorentz force density. The magnetic force separates into magnetic tension and magnetic pressure. It is coupled to the continuity equation and the ideal magnetohydrodynamic induction equation.

For smooth inviscid barotropic fluid motion without gravity or magnetic forces, let $w$ be the specific enthalpy with $\nabla w={\rho }^{-1}\nabla p$, and set $q=\mathbf{u}\cdot \mathbf{\omega }$. ThenThis follows by dotting the barotropic vorticity transport equation with $\mathbf{u}$ and using the divergence of a cross product. A volume integral is constant when the resulting boundary flux vanishes.

In Gaussian units, the equation in the title uses the gyrotropic pressure tensor $\mathbf{P}=p_{\perp }\mathbf{I}+(p_{\parallel }-p_{\perp })\mathbf{bb}$. This allows pressure anisotropy. Its magnetic force contains both magnetic pressure and magnetic tension.

The magnetic part of the Lorentz force is perpendicular to the particle velocity. Its power is therefore zero, so with no other forces the kinetic energy remains constant. This remains true relativistically because $dE/dt=\mathbf{F}\cdot \mathbf{v}$; a magnetic force can change momentum direction without changing total energy.

For constant mass density $\rho$, kinematic viscosity $\nu$ and magnetic permeability ${\mu }_0$, the incompressible momentum equation isIt is the Navier-Stokes equation with Lorentz force density. The magnetic force decomposes into magnetic tension and a magnetic pressure gradient. A uniform streamwise pressure gradient need not make ordinary pressure independent of transverse coordinates.

With the Einstein summation convention, the magnetic force has components $F_i=q{ϵ}_{ijk}{v}_j{B}_k$. Thus its matrix acting on velocity isThe Levi-Civita symbol changes sign on exchanging $i,j$, so $M^T=-M$; the diagonal vanishes and the matrix has zero trace. For nonzero $\mathbf{B}$, $M\mathbf{B}=q\mathbf{B}\times \mathbf{B}=0$, exhibiting a nontrivial kernel and proving noninvertibility. If $\mathbf{B}=0$ or $q=0$, $M$ is the zero matrix and is again noninvertible.

Equivalently, an odd-dimensional real antisymmetric matrix has $\det M=\det M^T=\det (-M)=-\det M$, hence zero determinant. The same antisymmetry gives ${\mathbf{v}}^{T}M\mathbf{v}=0$, expressing that the magnetic force does no work.

For one wire, Ampère's circuital law on a concentric circle of radius $r$ givesThe other wire experiences the Lorentz force on its electric current, $d\mathbf{F}=Id\mathbf{\ell }\times \mathbf{B}$. Its direction is perpendicular to this magnetic field, so the force per unit length isThe right-hand rule gives attraction for parallel electric currents in the same direction and repulsion for opposite directions.

For the upper wire, the upward magnetic force per unit length balances its weight per unit length, $\lambda g$, where $g$ is the gravitational acceleration. ThereforeThis assumes the lower wire is held fixed and both wires remain straight. Within the vertical-motion model, increasing the separation weakens the upward magnetic force, so this equilibrium is vertically restoring.

The component of magnetic force along the slope has magnitude $IBs\cos \varphi$. The rate at which the sliding bar does work against this magnetic braking isBy Ohm's law, $\mathcal{E}=IR$, soThis exactly equals the Joule heating in the resistor in the negligible-self-inductance approximation. The normal force does no work because it is perpendicular to the velocity.

The gravitational power is instead $mgv\sin \varphi$. By the work-energy theorem,It equals the Joule heating only at constant speed. For a concrete counterexample, during the initial acceleration the kinetic energy increases, so some gravitational power remains mechanical.

Use the coordinates in 8. At downhill speed $v$, the moving bar has $\mathbf{v}\times \mathbf{B}=-Bv\cos \varphi \hat{\mathbf{b}}$. The electromotive force and induced electric current are thereforeHere $I$ is positive along $-\hat{\mathbf{b}}$ across the sliding bar, then returns through the rails and resistor. This direction makes the magnetic force oppose downhill motion, as required by Lenz's law. The factor $\cos \varphi$ is essential: only the component of magnetic field normal to the rail plane threads the circuit. Rail and bar electrical resistances and circuit self-inductance are neglected relative to the stated large $R$ model.

Faraday's law of induction states that the oriented electromotive force around a closed material circuit is minus the rate of change of its linked magnetic flux:The surface orientation and circuit direction obey the right-hand rule. The minus sign expresses Lenz's law: the induced electric current, when allowed to flow, produces a magnetic field opposing the change of linked magnetic flux. For a stationary circuit this reduces to $\oint \mathbf{E}\cdot d\mathbf{\ell }=-d{\Phi }_B/dt$.

The source's “Lorenz force” is the Lorentz force, $q(\mathbf{E}+\mathbf{v}\times \mathbf{B})$. A wire element of cross-section $A_c$ contains $n{A}_{c}d\ell$ carriers of charge $q$ with drift velocity ${\mathbf{v}}_d$. Adding their magnetic forces and using $I=nq{A}_c{v}_d$ givesfor a straight wire in a uniform magnetic field. The length vector follows conventional electric current, irrespective of the carrier sign.

For the unlabelled upward-motion request, define $\hat{\mathbf{t}}=(\cos \varphi ,0,-\sin \varphi )$ down the slope, $\hat{\mathbf{b}}=(0,1,0)$ across the rails and $\hat{\mathbf{n}}=(\sin \varphi ,0,\cos \varphi )$ out of their plane. Choose the vertical magnetic field as $\mathbf{B}=B\hat{\mathbf{z}}$. An electric current $I$ across the bar in direction $-\hat{\mathbf{b}}$ gives a horizontal magnetic force $-IBs\hat{\mathbf{x}}$. The other forces are weight $-mg\hat{\mathbf{z}}$ and the rail normal force $N\hat{\mathbf{n}}$. Uniform upward velocity requiresConsequentlyEach symmetric rail supplies half of $N$. The electric current must make $I\mathbf{L}\times \mathbf{B}$ point horizontally toward the uphill side; for the coordinates just defined it flows along $-\hat{\mathbf{b}}$. Reversing the vertical magnetic field reverses this current direction. The external supply must maintain that electric current against the motional electromotive force during ascent. For $\cos \varphi =0$, no finite horizontal magnetic force can support uniform motion along the vertical rails.

Faraday's law for a moving circuit is $\mathcal{E}=\oint (\mathbf{E}+\mathbf{v}\times \mathbf{B})\cdot d\mathbf{\ell }=-d{\Phi }_B/dt$. The magnetic flux ${\Phi }_B$ is through a surface bounded by the instantaneous loop, with consistent orientation. For the stationary external field here, the motion supplies the motional electromotive force.

Let $x$ be downward penetration of the bottom segment, $v=\dot{x}$, and $K=B_{0}a$. While $0<x<b$, flux grows at rate $Kv$. Define positive current as the direction whose magnetic force is upward. With negligible self-inductance, $I=Kv/R$ andThe resistive magnetic entry of a falling loop givesThe initial acceleration is $g$ and the speed approaches $g\tau$ if the loop remains partly immersed long enough. Joule heating $I^{2}R$ removes gravitational energy. At the first $x=b$, both horizontal segments are inside the uniform field: the net flux becomes constant, the net magnetic force vanishes, and this zero-inductance model has $I=0$ thereafter. The subsequent speed increases with acceleration $g$.

For negligible electrical resistance, conservation of total linked flux instead gives $LI=Kx$ in the braking-current convention. Thereforeand the lossless limit of magnetic loop braking yieldsHere energy is stored reversibly in $L{I}^2/2$, not dissipated. If $b>2g/{\omega }^2$, the loop reverses before complete entry and executes the ideal oscillatory motion while straddling the boundary. If $b<2g/{\omega }^2$, it reaches complete entry on its first descent; thereafter the current is constant, the forces on its two horizontal segments cancel, and it falls with acceleration $g$. The exact grazing case needs care at the ideal discontinuous field boundary. An entry-only sinusoid must not be extrapolated through full immersion.

Take $\mathbf{B}=B\hat{\mathbf{z}}$. The magnetic Lorentz force givesThus $v_x+i{v}_y=(v_x(0)+i{v}_y(0))e^{-i\Omega t}$, while $v_z$ is constant. Integrating the transverse velocity gives a circle of radius $r_L=v_{\perp }/|\Omega |$ about a fixed guiding centre; adding $z(t)=z_0+v_{\parallel }t$ gives a helical motion in a uniform magnetic field. The magnetic force does no work, so the speed and kinetic energy remain constant. The handedness reverses with the sign of $q$. If $v_{\parallel }=0$ the helix degenerates to a circle; if $v_{\perp }=0$ it is a straight line parallel to the field.

An RF cavity stores an oscillating electromagnetic field. Its longitudinal electric field does work on a correctly phased bunch, with energy gain $\Delta E=q{V}_{\mathrm{RF}}\mathcal{T}\sin {\varphi }_s$ for a chosen phase convention and transit-time factor $\mathcal{T}$. The bunch arrival time is synchronized so successive gaps accelerate rather than decelerate it. The magnetic force guides and focuses the beam but does no work.

A linear accelerator uses new accelerating structures along its path; cell lengths and timing accommodate the particle speed, and each bunch normally traverses a given section once. A circular synchrotron reuses the same RF cavities each revolution. Its synchronous phase supplies both acceleration and radiation replacement; RF frequency or harmonic synchronization and bending-field strength follow the required orbit and particle speed. Longitudinal phase stability captures nearby particles into RF buckets.

For very high-energy electron and positron colliders, a linear geometry avoids repeated synchrotron losses: at fixed radius, loss per turn scales as $E^4/m^4$. This is not a statement that all high-energy colliders are preferably linear. The LHC was already a circular proton collider in the 2015 context; the proton's much larger mass greatly suppresses radiation at the same energy. Circular electron machines can also be competitive at lower energies. Magnet limits, luminosity, cost and beamstrahlung enter any real comparison.

The high-pressure limit of the slow magnetosonic wave approaches an incompressible polarization in the plane of the background magnetic field and wavevector, perpendicular to the latter. It has the displayed leading dispersion relation, the same as the out-of-plane Alfvén wave. Magnetic tension is the restoring force, and pressure balances the compressive component of the magnetic force. In exactly incompressible homogeneous ideal magnetohydrodynamics it is one of the two transverse magnetic polarizations.

A rectangular loop entering a uniform magnetic field while only one horizontal segment is immersed has motional electromotive force $Bav$. Neglecting self-inductance, its electric current is $Bav/R$ and its upward magnetic force is $(Ba{)}^{2}v/R$. The equation $m\dot{v}=mg-(Ba{)}^{2}v/R$ gives exponential relaxation toward $mgR/(Ba{)}^2$. This entry model ends when both horizontal segments enter the field; then their net magnetic force and the rate of change of external magnetic flux vanish.