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Write the wire diameter as $d_w$ to distinguish it from other geometric labels. Its cross-sectional area is $A_w=\pi d_w^2/4$, and the total wire length is $4a$. From the electrical resistivity, its electrical resistance isWhile the loop straddles the boundary, let $h$ be the height of each vertical side remaining in the magnetic field. The linked magnetic flux has magnitude $Bah$, with $\dot{h}=-v$. Faraday's law therefore gives $|\mathcal{E}|=Bav$. Neglecting the loop's self-inductance, Ohm's law givesThe into-page magnetic flux is decreasing, so Lenz's law requires a clockwise electric current as viewed in the diagram. If the loop is wholly inside a uniform magnetic field, or wholly outside it, its magnetic flux is constant and this induced electromotive force is zero. During a rapid transient one should instead solve $L\dot{I}+RI=Bav$; at the assumed steady speed the same boxed relation holds.

Use $d\mathbf{F}=Id\mathbf{\ell }\times \mathbf{B}$ for the magnetic Lorentz force on each wire segment. Let $+x$ point right, $+y$ upward and $+z$ out of the page, so $\mathbf{B}=-B\hat{\mathbf{z}}$. With clockwise electric current, the four magnetic forces areThe upper side carries electric current to the right and experiences an upward force, opposing the fall. The bottom side lies outside the magnetic field and has no magnetic force. On the right side the electric current is downward and the force is outward to the right; on the left it is upward and the force is outward to the left. Each side's in-field vertical length is $h$, so these horizontal forces cancel in the net force on the loop.

The wire's total mass is $M=4a{\rho }_m{A}_w$. Uniform wire gives each side a downward weight $Mg/4$. Internal mechanical forces at the corners transmit the unequal loads and maintain the assumed square shape; they cancel for the loop as a whole. The total external vertical force is therefore $Mg-IaB$ downward.

At terminal velocity, the upward magnetic force balances the weight. Substitute the electric current from the preceding solution:The geometric factors cancel, givingHere $g=9.81\mathrm{m}{\mathrm{s}}^{-2}$, $B=1\mathrm{T}$, ${\rho }_m=8960\mathrm{kg}{\mathrm{m}}^{-3}$ and ${\rho }_c=1.7\times 10^{-8}\mathrm{\Omega }\mathrm{m}$. As numerical checks, the specified $a=0.08\mathrm{m}$ and $d_w=0.001\mathrm{m}$ give $M=2.252\times 10^{-3}\mathrm{kg}$, $R=6.926\times 10^{-3}\mathrm{\Omega }$ and $I_s=0.276\mathrm{A}$.

The gravitational power lost, $Mg{v}_s$, equals the Joule heating $I_s^{2}R$, consistent with energy conservation. This terminal velocity applies only while the top arm remains in the magnetic field; after it exits, the magnetic braking ceases, apart from any brief inductive transient.

Choose the entry point as $(0,0)$, the initial velocity as $v\hat{\mathbf{x}}$, and the slit as $(d,a)$, with $a,d>0$. A magnetic field perpendicular to the page changes the direction of the velocity, while the Lorentz force does no work. For positive $q$, an upward initial force requires the magnetic flux density to point into the page. The resulting circular motion has centre $(0,r)$, since its initial tangent is horizontal.

The slit must lie on the circle, giving $d^2+(a-r{)}^2=r^2$ and hence $r=(d^2+a^2)/(2a)$. Equating the centripetal force to the magnetic part of the Lorentz force, $m{v}^2/r=|q|vB$, givesA negative electric charge requires the opposite direction of magnetic flux density. These expressions assume nonrelativistic motion and negligible energy loss through electromagnetic radiation.

For an opaque screen, reaching the slit must be the first encounter with $x=d$. The velocity there has $v_x/v=(d^2-a^2)/(d^2+a^2)$, so passage from left to right requires $a<d$, as in the drawn arrangement. At $a=d$ the trajectory merely touches the screen. If $a>d$, the same circle first meets the screen at height $d^2/a<a$, where the particle is blocked; its later passage through $(d,a)$ does not solve the opaque-screen problem. This qualification distinguishes a geometric circle through two points from an accessible slit trajectory.

Assume a passive linear electrical circuit in sinusoidal steady state, with transients absent and $V_0$ the peak voltage. Write $Z=|Z|e^{i\varphi }=R+iX$. The electric current is $I(t)=I_0\cos (\omega t-\varphi )$, with $I_0=V_0/|Z|$. Average the instantaneous electric power $V(t)I(t)$ over a full period. The double-frequency term averages to zero, leavingThereforeEquivalently, root mean square values give $⟨P⟩=V_{\mathrm{rms}}{I}_{\mathrm{rms}}\cos \varphi$. The reactance exchanges stored electric field or magnetic field energy with the source, whereas the real part produces Joule heating. A purely imaginary complex impedance dissipates no mean electric power under these ideal assumptions.

The electric field does work $qEd$ on the initially stationary particle. Thus the entry speed into the magnetic region isIn the source geometry the magnetic field is perpendicular to the page, and the particle enters perpendicular to the straight interface. Its return to that interface occurs after a semicircle. The separation between exit and return points is its orbit diameter, so the required separation $d$ gives $r=d/2$. ConsequentlyFor the drawn field into the page and a positive charge entering upwards, the trajectory bends left. Its speed is constant while in the magnetic field, so the time there is the semicircle length divided by $v_0$:The last form is half the cyclotron period and independently checks the geometric calculation.

The Lorentz force isFor $\mathbf{E}=0$, the magnetic part is perpendicular to the velocity, so its instantaneous power $\mathbf{F}\cdot \mathbf{v}$ vanishes. The speed and kinetic energy therefore remain constant. Decompose the initial velocity into $v_{\parallel }$ along the uniform magnetic field and $v_{\perp }$ perpendicular to it. The parallel component is unchanged, while the perpendicular component rotates uniformly.

For a nonrelativistic particle with $qB\ne 0$, the resulting helical motion hasHere $r_L$ is the Larmor radius, ${\omega }_c$ the cyclotron frequency in angular units, and $h$ the axial advance per turn. Opposite signs of electric charge give opposite senses of rotation. Pure perpendicular motion gives a circle; pure parallel motion gives a straight line. If $q=0$ or $B=0$, there is no magnetic deflection. At relativistic speeds, replace $m$ in the radius and period by $\gamma m$; $\gamma$ stays constant because the magnetic force does no work.

Let $A=\pi r^2$ and choose the rotating area normal to coincide initially with $+z$. At time $t$ its angle to the applied magnetic field is $\omega t$, so the external magnetic flux is ${\Phi }_B=BA\cos \omega t$. Faraday's law therefore givesIn the high-electrical resistance approximation, neglecting the loop's self-inductance, Ohm's law givesThe displayed signs specify positive circulation by the chosen normal; reversing the reference circulation reverses both signs. The high-electrical resistance assumption is quantitatively $\omega L\ll R$ if the loop has self-inductance $L$. Otherwise the current lags the induced electromotive force and must be obtained from $L\dot{I}+RI=\mathcal{E}$.

For a steady uniform current density in a nonmagnetic wire, the enclosed electric current at radius $r<a$ is $I_{\mathrm{enc}}=I{r}^2/a^2$. A circular Amperian loop and Ampère's law then giveThe magnetic field circles the wire according to the right-hand rule relative to the current direction. Its magnitude increases linearly from zero on the axis. The uniform-current assumption excludes skin effect.

For a long air-cored solenoid, Ampère's law applied to a rectangular loop running inside and outside the winding gives $B\ell \simeq {\mu }_{0}NI$, with the external magnetic field negligible away from the ends. Hence $B={\mu }_{0}nI$ along the winding axis, where $n=N/\ell$. Each turn links magnetic flux $BA$ with $A=\pi a^2$, so the total magnetic flux linkage is $NBA=LI$. It follows thatHere $n=2000{\mathrm{m}}^{-1}$, $a=0.010\mathrm{m}$ and $\ell =0.50\mathrm{m}$, so $N=1000$ andThe value uses the stated long-solenoid approximation and ignores end corrections and winding resistance.

In source-free vacuum, Maxwell's equations readTake the curl of the third equation and use the curl of the curl identity $\nabla \times (\nabla \times \mathbf{E})=\nabla (\nabla \cdot \mathbf{E})-{\nabla }^2\mathbf{E}$. ThenThe prescribed plane wave travels in the $+z$ direction. Its polarization is along $x$, and Faraday's law requires $B_y=(k/\omega )E_x$ with $\omega =ck$. The wave's magnetic flux density is thereforeThe direction is also fixed by $\mathbf{E}\times \mathbf{B}$ pointing along the propagation direction. A superposed static uniform magnetic field is not fixed by the oscillatory electric field; the expression above is the magnetic field belonging to the wave itself.

For a current element $Id\mathbf{\ell }$, the Biot-Savart law isTake a straight wire along $z$, extending from $-a$ to $a$, and observe its magnetic field at $(s,0,0)$. All contributions point in $+\hat{\mathbf{\varphi }}$ for positive $I$ along $+z$, by the right-hand rule. Their magnitude integrates toFor $a/s\to \infty$ this becomes ${\mu }_{0}I/(2\pi s)$, so $\oint \mathbf{B}\cdot d\mathbf{\ell }=2\pi s{B}_{\varphi }={\mu }_{0}I$, exactly as Ampère's law requires for an infinite straight wire.

Each long wire lies in the other's magnetic field, $B={\mu }_{0}I/(2\pi d)$. The magnetic Lorentz force per unit length has magnitudeFor the two-wire circuit's self-inductance, a nonzero wire radius is essential. Introduce a radius $a_w\ll d$. Between the wires their magnetic fields add. The external magnetic flux linkage per unit length through the connecting ribbon is, to leading order in $a_w/d$,It follows that the leading thin-wire loop self-inductance per unit length isIt is the loop value for the pair, not the contribution of just one conductor. Corrections from the wire interior depend on the current distribution: for nonmagnetic round wires with uniform direct electric current, the pair contributes another ${\mu }_0/(4\pi )$; for a surface current the internal contribution is zero. The leading logarithm is common to both models.

TODO (source data): the source gives no wire radius or current-distribution convention. There is no finite radius-independent value of the requested self-inductance: taking ideal zero-width wires sends $L_{\mathrm{ext}}^{\prime }$ to infinity. The force and the parameterized leading self-inductance above are determined, but a unique finite answer requires that additional specification. For example, changing $a_w$ to $a_w/2$ changes the leading result by ${\mu }_0\ln 2/\pi$ even at fixed separation.

Put the three wires at $(x,y)=(-d,0),(0,0),(d,0)$ with $d=1\mathrm{cm}$ and each current along $+z$, out of the page. Adding their Biot-Savart law magnetic fields givesFor $y\ne 0$, all three $x$-components have the same sign, so they cannot cancel. On $y=0$, a zero of the remaining component satisfiesExcluding the wire positions themselves, the two and only two finite zeros areThe magnetic field lines circulate counterclockwise about each current near that wire. Far away they circle the combined total current $3I$. The two nulls separate the nearby circulation regions in the original magnetic field line plot below.

In a source-free vacuum, the Maxwell equations in SI arewhere ${ϵ}_0$ and ${\mu }_0$ are the vacuum permittivity and vacuum permeability. Take the curl of the last equation, substitute Faraday's law of induction, and use the curl of the curl identity:The zero divergence therefore gives the wave equationFor a propagating plane wave, write $\mathbf{B}=\mathrm{Re}[{\mathbf{B}}_0{e}^{i(\mathbf{k}\cdot \mathbf{r}-\omega t)}]$. The wave equation gives ${\omega }^2=c^2|\mathbf{k}{|}^2$, and Gauss's law for magnetism gives $\mathbf{k}\cdot {\mathbf{B}}_0=0$. Thus the oscillating magnetic field is a transverse wave: it is perpendicular to the propagation wavevector. A static background magnetic field, if present, is separate from this propagating component.

The current density is the moving charge density times its velocity:For a cylindrical Gaussian surface of radius $r<R$ and length $\ell$, cylindrical symmetry makes the electric field radial. Gauss's law gives $E(2\pi r\ell )=\rho \pi r^2\ell /{ϵ}_0$. A circular Ampère's circuital law contour encloses electric current $I(r)=\rho v_0\pi r^2$, giving $B(2\pi r)={\mu }_{0}I(r)$. HenceThe right-hand rule fixes the positive azimuthal direction of the magnetic field for positive electric charge moving along $+z$. Both the electric field and the magnetic field vanish on the axis. These are the steady long-cylinder electric field and magnetic field; end effects and changes of the prescribed beam distribution are excluded.

For a steady filamentary electric current $I$, the Biot-Savart law isHere $\mathbf{r}$ is the observation position, ${\mathbf{r}}^{\prime }$ the source position, $d{\mathbf{\ell }}^{\prime }$ points in the direction of positive electric current, and ${\mu }_0$ is the vacuum permeability. A circular loop has centre magnetic field ${\mu }_{0}I/(2r)$: the cross product has magnitude $rd{\ell }^{\prime }$ and a constant axial direction, and $\int d{\ell }^{\prime }=2\pi r$.

Introduce the constant signed angular velocity $\Omega$ about the positive axis. A ring's electric charge $dq=2\pi \sigma rdr$ crosses any radial cut once per rotation, so its electric current is $dI=\Omega dq/(2\pi )=\sigma \Omega rdr$. Adding loop contributions yields the magnetic field of a rotating charged annulus:The direction follows the right-hand rule for positive electric charge; reversing either $Q$ or $\Omega$ reverses the magnetic field. The source supplies no rotation rate, so the answer necessarily retains $\Omega$.

For steady currents in vacuum, Ampère's circuital law isHere $C$ is an oriented closed curve, $d\ell$ its tangent line element, and $I_{\mathrm{enc}}$ the signed electric current through a spanning surface oriented by the right-hand rule. A circular contour about a straight infinite wire has constant tangential magnetic field, so $B(r)={\mu }_{0}I/(2\pi r)$. The Lorentz force on a length $L$ of the other wire is $IL\hat{\ell }\times B$. ThereforeThe right-hand rule makes the force on each parallel, same-direction current point toward the other wire.

Take an electric current $I_1$ in the long solenoid. Away from its ends, Ampère's circuital law gives the approximately uniform magnetic fieldHere ${\mu }_0$ is the vacuum permeability; an air core is understood. Each turn of the short surrounding coil encloses the magnetic flux $B_1\pi R^2$. Thus its magnetic flux linkage is ${\Lambda }_2=N_2{B}_1\pi R^2$, and the mutual inductance isThe short coil lies in the nearly uniform central magnetic field; its own axial length therefore does not enter. With consistent winding orientations, Faraday's law of induction gives ${\mathcal{E}}_2=-Md{I}_1/dt$. Reversing a winding reverses this signed coupling, while the stated magnitude is unchanged.

Let $\theta =\omega t+{\theta }_0$ be the angle between the coil's area normal and the magnetic field. Its magnetic flux linkage is $\Lambda =NB\pi r^2\cos \theta$. Consequently Faraday's law of induction givesThe sign refers to the winding orientation used to define the normal. The initial phase ${\theta }_0$ depends on the chosen time origin; one can set it to zero when the normal first points along $+x$. The minimum absolute value is zero, attained at $\theta =m\pi$: the coil's normal is parallel or antiparallel to the $x$-axis, and the plane of the coil is perpendicular to that axis. At this orientation the magnetic flux is extremal and its derivative with respect to time vanishes.