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For a smoothly moving circuit and a divergence-free magnetic field, the magnetic flux derivative is $d\Phi /dt={\int }_S{\partial }_t\mathbf{B}\cdot d\mathbf{S}-{\oint }_C(\mathbf{v}\times \mathbf{B})\cdot d\mathbf{r}$. The minus sign follows by orienting the swept side surface compatibly with the two spanning surfaces. Combining this transport identity with Faraday's law yields the moving-loop electromotive-force law.

Motional electromotive force is the contribution caused by a conductor moving through a magnetic field. The magnetic part of the Lorentz force separates charges within the conductor. For a bar of length $L$ moving at velocity $v$ perpendicular to a uniform field, its magnitude is $BL|v|$; the sign follows the chosen circuit orientation. Together with the electric-field contribution it enters Faraday's law for moving circuits.

Use compatible orientations for the loop and spanning surface. The relevant Maxwell equations areThe magnetic flux transport for a moving circuit separates the change of the field from the motion of its boundary. To see the boundary term, freeze $\mathbf{B}$ at time $t$. Orient the closed swept surface as $S(t+\delta t)-S(t)+S^{\prime }$. Its side element isto first order. Hence its flux is $-\delta t{\oint }_C\mathbf{B}\cdot (\mathbf{v}\times d\mathbf{r})$. By the divergence theorem and $\nabla \cdot \mathbf{B}=0$, the sum of the three fluxes is zero, giving the purely geometric changeAdding the temporal field change and applying Stokes theorem yieldsThusThe boundary contribution is the motional electromotive force; the field contribution is Faraday's law.

Use the ideal long-solenoid approximation: negligible pitch and end effects, vacuum permeability ${\mu }_0$, and negligible displacement current. By cylindrical symmetry the magnetic field is axial. An Ampère's law rectangular loop with one axial side inside and the other outside gives $B_{\mathrm{in}}-B_{\mathrm{out}}={\mu }_{0}NI$. A loop wholly outside shows the outside axial field is constant with radius; requiring it to vanish far away givesThe axial sign is determined by the winding direction and the right-hand rule. The small axial component of current from the spiral pitch is neglected in this idealization.

There are $Nl$ turns, and the magnetic flux through each is $\pi a^2{\mu }_{0}NI$. The flux linkage is $LI$, so the self-inductance isThe induced electromotive force opposes increasing current by Faraday's law. Together with Ohm's law, the circuit equation is $L\dot{I}+RI={\mathcal{E}}_0$. Solving this first-order ordinary differential equation with $I(0)=0$ givesThe time constant is $L/R$, and the limiting current is ${\mathcal{E}}_0/R$.

For a moving closed circuit $C(t)$ bounding an oriented surface $S(t)$, Faraday's law isHere $\mathbf{E}$ is the electric field, $\mathbf{B}$ the magnetic field, $\mathbf{v}$ the local velocity of a circuit element, and ${\mathcal{E}}_{\mathrm{em}}$ the electromagnetic electromotive force per unit charge around the circuit. The line orientation and normal $\mathbf{n}$ obey the right-hand rule. The flux derivative is a total derivative and includes both changing field and changing surface. The term $\mathbf{v}\times \mathbf{B}$ is the motional electromotive force. A battery contributes an additional non-electromagnetic source electromotive force.

In a charge-free and current-free vacuum, Maxwell equations giveTake the curl of Faraday's law and use the curl of the curl identity $\nabla \times (\nabla \times \mathbf{E})=\nabla (\nabla \cdot \mathbf{E})-{\nabla }^2\mathbf{E}$. The divergence term vanishes by Gauss's law, while the time derivative of the magnetic curl supplies $-{\mu }_0{ϵ}_0{\partial }_t^2\mathbf{E}$. ThusThis is the transverse electromagnetic wave equation.

For the oscillatory magnetic field, Faraday's law gives $\mathbf{B}=(1/\omega )\mathbf{k}\times \mathbf{E}$. Indeed its time derivative equals $-\nabla \times \mathbf{E}$, and the dispersion relation makes it satisfy the remaining Maxwell equations. ThusFor $\omega =ck$, its amplitude is $E_0/c$. The magnetic field, electric field and direction of propagation are mutually perpendicular. A constant background magnetic field could be added; the expression above is the wave-associated field.

Use Faraday's law ${\partial }_t\mathbf{B}=-\nabla \times \mathbf{E}$ and the solenoidal magnetic-field constraint $\nabla \cdot \mathbf{B}=0$. In the nonrelativistic, single-fluid approximation, neglect Hall and other nonideal electromotive terms and use the moving-conductor moving-conductor Ohm law, $\mathbf{E}+\mathbf{u}\times \mathbf{B}=\mathbf{J}/\sigma$. Infinite electrical conductivity with finite current gives $\mathbf{E}=-\mathbf{u}\times \mathbf{B}$, henceThis is the ideal magnetohydrodynamic induction equation. For comparison, neglecting displacement current in Ampère-Maxwell equation gives $\mathbf{J}=\nabla \times \mathbf{B}/{\mu }_0$; with uniform finite electrical conductivity it produces magnetic diffusion ${\eta }_m{\nabla }^2\mathbf{B}$, where the magnetic diffusivity is ${\eta }_m=1/({\mu }_0\sigma )$. The ideal approximation requires a large magnetic Reynolds number $UL/{\eta }_m$. Dropping displacement current is useful for this finite-conductivity comparison, but Faraday's law and the ideal Ohm relation already suffice for the ideal induction equation.

Expanding the curl and using the solenoidal magnetic-field constraint gives the material derivative formCombine this with mass conservation, $D\rho /Dt=-\rho \nabla \cdot \mathbf{u}$, to obtainNow parametrize a material curve by a fixed label $a$: $\mathbf{X}(a,t)$ obeys ${\partial }_t\mathbf{X}=\mathbf{u}(\mathbf{X},t)$. Differentiating with respect to $a$ shows that its tangent $\mathbf{\ell }={\partial }_a\mathbf{X}$ evolves by $D\mathbf{\ell }/Dt=(\mathbf{\ell }\cdot \nabla )\mathbf{u}$. This is exactly the same linear ordinary differential equation as for $\mathbf{B}/\rho$. Initially parallel tangents remain parallel by uniqueness, with a label-dependent proportionality factor constant along each particle trajectory. Thus magnetic field lines are transported as material curves, wherever the field and fluid flow are smooth and the field is nonzero. This is the field-line form of magnetic flux freezing.

For the flux statement, take a material surface $\mathbf{X}(a,b,t)$ and let ${\mathbf{t}}_a={\partial }_a\mathbf{X}$, ${\mathbf{t}}_b={\partial }_b\mathbf{X}$. Both tangents obey $D{t}_i/Dt=({\partial }_j{u}_i)t_j$. Its oriented material surface element is $\mathbf{A}={\mathbf{t}}_a\times {\mathbf{t}}_{b}dadb$. Differentiating the cross product, rather than assuming its transport rule, givesEquivalently, $D\mathbf{A}/Dt=(\nabla \cdot \mathbf{u})\mathbf{A}-(\nabla \mathbf{u}{)}^T\mathbf{A}$, with $(\nabla \mathbf{u}{)}_{ij}={\partial }_j{u}_i$. Contracting this derived rule with the induction equation gives a pointwise cancellation:Integrating over the fixed material labels therefore proves conservation of flux through an open material surface:The surface need not be closed; its boundary is carried with the fluid. The result follows from material transport, rather than from the zero flux through a closed surface.

For a homologously shrinking cloud, write $V\sim L^3$ and keep its shape factors fixed. Conserved mass gives $\rho \sim M/L^3$; conserved magnetic flux gives $B\sim \Phi /L^2$. Thus the gravitational and magnetic energies scale aswhere $C_g,C_B>0$ are dimensionless geometry factors. Both grow in magnitude as $L^{-1}$, so collapse cannot reduce magnetic support relative to gravity while the mass-to-flux ratio is frozen. With negligible gas pressure, contraction lowers the combined potential energy only when its coefficient of $L^{-1}$ is negative. Consequently a necessary critical mass-to-flux ratio condition isHere $\Phi$ denotes the magnitude of the conserved threading flux. The numerical coefficient depends on geometry and boundary conditions; the scaling argument does not determine it or make the condition sufficient in the presence of other support.

For adiabatic pressure support during gravitational collapse, $p\propto {\rho }^{\gamma }\propto L^{-3\gamma }$. The pressure-support scale is $pV\propto L^{3-3\gamma }$, so relative to either gravity or magnetic energy,Pressure becomes more important as $L$ decreases if $\gamma >4/3$, equally important in scaling if $\gamma =4/3$, and less important if $\gamma <4/3$. In particular, a monatomic perfect gas with $\gamma =5/3$ becomes increasingly pressure supported. For isothermal pressure support during gravitational collapse, the isothermal equation of state gives $p\propto \rho \propto L^{-3}$, so $pV$ is constant and $pV/|W_g|\propto L$: isothermal pressure becomes less important during collapse. The same comparisons hold against magnetic support because its energy has the same $L^{-1}$ scaling as gravity.

The Maxwell equations with vacuum permittivity and permeability give $\rho ={ϵ}_0\nabla \cdot E$ and $J={\mu }_0^{-1}\nabla \times B-{ϵ}_0{\partial }_{t}E$. The current identity is the Ampère-Maxwell equation, and Faraday's law gives ${\partial }_{t}B=-\nabla \times E$. With $g={ϵ}_{0}E\times B$,The identity $(\nabla \times B)\times B=(B\cdot \nabla )B-\frac{1}{2}\nabla B^2$, and its electric analogue, turn this into the divergence of the Maxwell stress tensor $T$. Thus the tensor in the question is its negative:The vector $g$ is electromagnetic momentum density, and ${\sigma }_{ij}$ is the outward flux of its $i$th component across a surface normal in the $j$ direction. In static conditions the total force on matter is $F_i=-\oint {\sigma }_{ij}{n}_{j}dS$; generally subtract the time derivative of the enclosed electromagnetic field momentum as well.

For the two infinite sheets, superposition and the electromagnetic field jumps give $E=(\sigma /{ϵ}_0)e_x$, $B={\mu }_{0}K{e}_y$ between them and zero outside. Substitution yields exactlyA thin pillbox enclosing the $x=0$ sheet has only its $+x$ face in the interior electromagnetic field. ThereforeDirectly, omit the sheet's self-electromagnetic field: the other sheet supplies $E_{\mathrm{other}}=\sigma e_x/(2{ϵ}_0)$ and $B_{\mathrm{other}}={\mu }_{0}K{e}_y/2$. The Lorentz force per area is $\sigma E_{\mathrm{other}}+K{e}_z\times B_{\mathrm{other}}$, giving the same result. The electric contribution is attractive and the magnetic contribution repulsive.

If $K=\sigma v$ for a single moving charge population, use ${\mu }_0{ϵ}_0=c^{-2}$ to obtain $F/A={\sigma }^2(1-v^2/c^2)e_x/(2{ϵ}_0)$. For physical $|v|<c$ this is attractive, so motion of these charges cannot make the force repulsive. Independently imposed currents need not satisfy that restriction on $K/\sigma$.

Apply the curl of the curl identity $\nabla \times (\nabla \times E)=\nabla (\nabla \cdot E)-{\nabla }^{2}E$ to Faraday's law, and use the Ampère-Maxwell equation. This givesApplying the same identity to $B$, using $\nabla \cdot B=0$, gives

For a spherically symmetric charge density, Gauss's law on a sphere of radius $r$ givesFor the unit-density ball of radius $a$,The potential is continuous at $r=a$ and tends to zero at infinity.

Let $\mathbf{p}=(1,0,0)$ and write $1_{S_n}$ for the indicator of $S_n$. The layered density decomposes asThe point $\mathbf{p}$ lies on every sphere $S_n$, whose radius is $R_n=4^{-n}$ and whose outward radial direction there is $\hat{\mathbf{x}}$. A uniform ball of density $c$ contributes $c{R}_n/3$ to the field and $c{R}_n^2/3$ to the potential at its surface. Thereforeand