With phasors proportional to $e^{i\omega t}$, the series RLC circuit hasFor fixed $V_0$ and frequency-independent positive $R$, the maximum occurs when the reactances cancel. ThusThis is the driven current resonance; a damped free-oscillation frequency is a different quantity.

After the electric field is removed, the uniform magnetic field preserves the speed. Decompose the instantaneous initial velocity as ${\mathbf{v}}_0={\mathbf{v}}_{\parallel }+{\mathbf{v}}_{\perp }$, with $\hat{\mathbf{b}}=\mathbf{B}/B$ and signed $\Omega =qB/m$. Thenand, integrating,This is a helical motion in a uniform magnetic field of radius $m|{\mathbf{v}}_{\perp }|/(|q|B)$ and angular frequency $|q|B/m$. It is a circle if ${\mathbf{v}}_{\parallel }=0$, and a straight line if ${\mathbf{v}}_{\perp }=0$. The formula assumes a uniform, nonzero field and nonrelativistic motion.

For the intended force-balanced straight-line motion, the Lorentz force vanishes:For a nonzero uniform magnetic field, this implies $\mathbf{E}\cdot \mathbf{B}=0$ andwhile a component parallel to $\mathbf{B}$ is unrestricted. A straight trajectory by itself states only that the force has no transverse component, $\mathbf{v}\times (\mathbf{E}+\mathbf{v}\times \mathbf{B})=0$. For example, parallel electric field and motion can give straight but accelerated motion. The boxed cancellation is the usual constant-velocity interpretation of the source.

The loop area is $A=\pi (5\times 10^{-3}{)}^2{\mathrm{m}}^2$. Integrating Ohm's law and Faraday's law through the whole decay givesThereforeThe switching time cancels: a faster linear flux decay produces a larger electric current for a proportionally shorter interval. If the coil's self-inductance is retained, integrating $LdI/dt$ gives zero provided the initial and final currents vanish and the stated total charge includes the decay tail.

Faraday's law states that the induced electromotive force equals minus the rate of change of linked magnetic flux:The minus sign expresses Lenz's law: the induced electric current acts to oppose the change in magnetic flux. For a fixed contour this is equivalent to $\oint \mathbf{E}\cdot d\mathbf{\ell }=-d\int \mathbf{B}\cdot d\mathbf{S}/dt$.

The phase $kz-\omega t$ describes propagation along $+z$. With $\psi =kz-\omega t$, the vacuum plane electromagnetic wave hasThus $B_x=-3A\sin \psi /c$, $B_y=2A\sin \psi /c$, and $B_z=0$. The field amplitudes are $E_0=\sqrt{13}|A|$ and $B_0=\sqrt{13}|A|/c$, givingThe instantaneous magnitude ratio also equals $c$ wherever the fields are nonzero; at a node both vanish together. Their Poynting vector points along $+z$, checking the relative component signs.

In charge- and current-free vacuum, $\nabla \cdot \mathbf{E}=\nabla \cdot \mathbf{B}=0$. Taking the curl of Faraday's law and using the Ampère-Maxwell equation givesSince $\nabla \times (\nabla \times \mathbf{E})=\nabla (\nabla \cdot \mathbf{E})-{\nabla }^2\mathbf{E}$, it follows thatSubstituting a plane wave with phase $\mathbf{k}\cdot \mathbf{r}-\omega t$ gives $k^2={\mu }_0{ϵ}_0{\omega }^2$ and thereforeThe divergence equations make the waves transverse, and the curl equations fix $\mathbf{B}=\hat{\mathbf{k}}\times \mathbf{E}/c$.

For a charging parallel-plate capacitor of area $A$, neglect fringing. The field in the gap is $E=Q/({ϵ}_{0}A)$, so the displacement current through the gap isAn Amperian loop around the feed wire can bound a surface cutting the wire, which has conduction current $I$, or a bulging surface passing through the dielectric gap, which has no conduction current. Including displacement current makes both surfaces give the same magnetic circulation. It is also required by charge conservation: the divergence of the Ampère-Maxwell equation and Gauss's law yields $\nabla \cdot \mathbf{J}+\partial \rho /\partial t=0$, the continuity equation.

Maxwell's equations in matter, with free charge and current densities, areFor vacuum, $\mathbf{D}={ϵ}_0\mathbf{E}$ and $\mathbf{H}=\mathbf{B}/{\mu }_0$, so the last equation is $\nabla \times \mathbf{B}={\mu }_0\mathbf{J}+{\mu }_0{ϵ}_0\partial \mathbf{E}/\partial t$.

Relative to the real positive source phasor ${\tilde{V}}_{AB}=V_0$,Thus $V_{XZ}(t)=(V_0/5)\cos (\omega t-75.52^{\circ })$ lags $V_{AB}$. Reversing the voltage ordering to $V_{ZX}$ would add $180^{\circ }$.

Put ${\tilde{V}}_B=0$ and ${\tilde{V}}_A=V_0$, with ${\tilde{V}}_{PQ}={\tilde{V}}_P-{\tilde{V}}_Q$. The source current is $V_0/(10R)$ and the parallel-network voltage is ${\tilde{V}}_X=4V_0/5$. HenceThe voltage from X to each branch midpoint is its resistor drop, soSince $|1\pm i\sqrt{15}|=4$,Equal amplitude does not mean equal phase.

The source diagram has $2R$ in series with two parallel branches: $R+L$ above and $R+C$ below. With $s=\sqrt{15}$, their complex impedances are $Z_{+}=R(1+is)$ and $Z_{-}=R(1-is)$. Their parallel combination isThusThe input electric current is in phase with $V_{AB}$ even though the two branch currents separately have nonzero, opposite phase shifts.

Use the phasor convention $v(t)=\mathrm{Re}(\tilde{V}{e}^{i\omega t})$. The complex impedance is $Z=\tilde{V}/\tilde{I}$ in sinusoidal steady state. From $v_R=Ri$, $i_C=Cd{v}_C/dt$, and $v_L=Ld{i}_L/dt$,For this convention an ideal inductor has voltage leading current by $\pi /2$, while an ideal capacitor has current leading voltage by $\pi /2$.

At $(x,0)$ the distances from the two centres are equal:By the electrostatic shell theorem, a uniformly charged spherical shell gives zero field inside and its point-charge field outside. For $x<x_1$ the point is inside both shells. For $x_1<x<x_2$ only the small, negatively charged shell contributes. For $x>x_2$ both contribute, soThe requested magnitude isIt jumps from zero to $k_{e}q/a^2$ at $x_1$, decreases to $k_{e}q/(4a^2)$ just before $x_2$, jumps down to $k_{e}q/(8a^2)$ just after $x_2$, and then falls as $x^{-3}$. The ideal surface charge makes the field discontinuous at each crossing, so the one-sided limits, rather than a unique value on the sheet, specify the graph. The far field is the equatorial field of an electric dipole with electric dipole moment $\mathbf{p}=-qa\hat{\mathbf{y}}$.

Put charges $+Q$ and $-Q$ on the concentric conductors. Gauss's law gives $E(r)=Q/(4\pi {ϵ}_0{r}^2)$ between them, $a<r<2a$. The potential difference isThe capacitance of this vacuum spherical capacitor is thereforeThe conductors are equipotentials; a uniform dielectric filling would replace ${ϵ}_0$ by its permittivity.

In matter, Gauss's law for free electric charge isThe area normal points out of $\mathcal{V}$. In vacuum $\mathbf{D}={ϵ}_0\mathbf{E}$, so these become $\nabla \cdot \mathbf{E}=\rho /{ϵ}_0$ and $\oint \mathbf{E}\cdot d\mathbf{S}=Q_{\mathrm{enclosed}}/{ϵ}_0$.

At the largest admitted orbit, $r=R$ and $m_p{v}_{\max }=eBR$. ConsequentlyUsing $e=1.602176634\times 10^{-19}\mathrm{C}$ and $m_p\simeq 1.67262\times 10^{-27}\mathrm{kg}$ with the specified $B$ and $R$ yieldsThis is about $0.160c$; the displayed results use the nonrelativistic approximation explicitly requested by the paper. The drive amplitude controls the energy gain per gap crossing and hence the number of turns, but the ideal orbit-radius limit is set by $B$ and $R$.

For a proton, the fundamental cyclotron resonance condition isA half-orbit takes $\pi m_p/(eB)$, during which the applied voltage reverses sign once. The proton then crosses the gap in the opposite direction and receives another accelerating impulse. The electric force in the gap, rather than the magnetic bending inside a dee, provides the kinetic energy. The injection phase must be chosen so the first crossing accelerates the proton.

Take the initial velocity perpendicular to a uniform magnetic field. The magnetic Lorentz force is perpendicular to the motion, has constant magnitude $|q|vB$, and supplies the centripetal force:The rotation sense is fixed by $q\mathbf{v}\times \mathbf{B}$. A component of velocity parallel to the field remains constant and produces a helical motion in a uniform magnetic field; a circle requires that component to vanish. The speed-independent cyclotron frequency here assumes nonrelativistic motion.

The Lorentz force law isIts rate of work is $dT/dt=\mathbf{F}\cdot \mathbf{v}=q\mathbf{E}\cdot \mathbf{v}$, because $\mathbf{v}\cdot (\mathbf{v}\times \mathbf{B})=0$. Thus, with $\mathbf{E}=0$, the kinetic energy cannot increase: a static magnetic field changes the direction of motion but does no work.