Faraday's law relates induced electromotive force to the rate of change of magnetic flux linkage. For a coil with $N$ turns and a consistent right-hand orientation,Flux can change through field variation or circuit motion. Lenz's law is the minus sign: the induced current's field opposes the flux change, consistent with energy conservation. It opposes the change rather than necessarily opposing the original field.

For a linear fixed-geometry inductor, $N{\Phi }_B=LI$ and the self-induced electromotive force is $-LdI/dt$. The passive terminal voltage drop is instead $V=LdI/dt$. The stored magnetic energy isInductance is measured in henries. Transformers use mutual inductance: changing one winding's current induces an electromotive force in another.

A field assigns a physical quantity to each point. Gravitational field is force per unit test mass; electric field is force per unit test charge. In a conservative, time-independent field,Write $U=m\Phi$ for gravitational potential $\Phi$, or $U=q\varphi$ for electric potential $\varphi$, giving $\mathbf{g}=-\nabla \Phi$ and $\mathbf{E}=-\nabla \varphi$. A point mass has $U=-GMm/r$ with zero at infinity, while two point charges have $U=Qq/(4\pi {ϵ}_{0}r)$. Only potential-energy differences matter; a constant can be added freely. The mechanical energy $K+U$ is conserved if no nonconservative forces do work. A magnetic Lorentz force $q\mathbf{v}\times \mathbf{B}$ does no work on a point charge and generally cannot be represented by a position-only scalar potential energy.

In magnetostatics Ampère's circuital law is $\oint \mathbf{B}\cdot d\mathbf{l}={\mu }_0{I}_{\mathrm{enc}}$. Its general vacuum version adds ${\mu }_0{ϵ}_{0}d{\Phi }_E/dt$ on the right. Faraday's law is $\mathcal{E}=\oint \mathbf{E}\cdot d\mathbf{l}=-d{\Phi }_B/dt$ for a fixed loop. A circular Amperian path around a long straight wire giveswith direction given by the right-hand rule. The force on a wire element is $d\mathbf{F}=Id\mathbf{l}\times \mathbf{B}$; its magnitude is $IBdl\sin \chi$ and its direction follows this cross product.

Choose the straight-wire current rightward and the square below it, as printed. Its positive magnetic field there is into the page. With an into-page loop normal, the magnetic flux linkage isHere $\mathcal{M}$ denotes mutual inductance, not a mass. Choose positive coil current to produce an out-of-page field. Then its induced current in the resistance-dominated approximation isThis reproduces the specified sign with the declared orientation. It neglects coil self-inductance, requiring $\omega L_{\mathrm{coil}}\ll R$, as implicit in the requested expression.

The nearest side contains N parallel current segments and is at a uniform distance $s$, so its force magnitude isThe two sides perpendicular to the wire have opposite currents and equal magnetic fields at corresponding distances; their forces cancel pairwise. The far side opposes the near-side force and has the smaller field. ThusWhen $I$ and $i$ have the same sign under the chosen conventions, the near-side coil current opposes the straight-wire current and the total force is away from the wire; otherwise it is toward it. In particular the signed force alternates with $\sin (\omega t)\cos (\omega t)$. The coil is fixed, so these forces do no mechanical work on it. The mean Joule heating is

Substitution into the wave equation gives ${\partial }_x^2\psi +{\partial }_y^2\psi =-(k_x^2+k_y^2)\psi$ and ${\partial }_t^2\psi =-{\omega }^2\psi$. Therefore ${\omega }^2=c^2(k_x^2+k_y^2)$. The fixed edges at zero are automatically satisfied by the sine factors; the opposite edges requireNeither integer may vanish, because that would make the entire displacement zero.

The element's mass is $\rho dxdy$, and its speed is $-\omega A\sin (k_{x}x)\sin (k_{y}y)\sin (\omega t)$. Squaring and multiplying by half its mass gives the requested energy density. The two sine-squared spatial integrals are $a/2$ and $b/2$, soThere is no dissipation. The total mechanical energy equals the maximum kinetic energy, givingEquivalently, integrate the small-slope tension energy $\rho c^2|\nabla \psi {|}^2/2$ and use the dispersion relation.

For $b=4a/5$, the two lowest modes are $(p,q)=(1,1)$ and $(2,1)$:The fundamental has one antinode at $(a/2,b/2)$ and no interior nodal line. The next mode has a node along $x=a/2$ and opposite-sign antinodes at $(a/4,b/2)$ and $(3a/4,b/2)$. Every boundary is a node.

The square fundamental has $f=c\sqrt{2}/(2\beta a)$. Equating it to $f_{11}$ givesAt the same peak displacement amplitude and frequency, time-averaged kinetic energy is proportional to membrane area. ThereforeThe reciprocal ratio is $41/40$ if the rectangle is placed first.

Across a uniformly illuminated slit, the phase difference between its edges is $2\pi a\sin \theta /\lambda$. Integrating the complex amplitude across the slit gives the single-slit diffraction factorThus $\beta =\pi a/\lambda$ in the small-angle amplitude expression. The first noncentral zero has $q=\pi$, so ${\theta }_0\simeq \lambda /a$. At the stated approximate subsidiary maximum,The exact first side maximum solves $\tan q=q$, giving $q\simeq 4.493$ and intensity ratio $0.0472$; the requested approximation need not be mistaken for the exact maximum.

A virtual object is a point toward which incident rays are already converging and would meet if the lens were removed. With real-is-positive object/image distances it has $u<0$. The thin-lens equation gives $1/v=1/f-1/u$; as $u\to -\infty$, $v\to f$, so a distant virtual object images in the rear focal plane.

Without the lens, $y\simeq L\theta$ puts the central maximum at zero and the first dark fringes at $y=\pm L{\lambda }_0/a=\pm 0.100\mathrm{m}$. Subsequent zeros are integer multiples, and weaker subsidiary maxima lie between them.

For the finite-distance pattern compression, take the old screen as the prospective image plane at $L=6\mathrm{m}$, with $u=-L$, and the new screen at $d=0.10\mathrm{m}$. ThenThe plane surface has infinite radius of curvature. The transverse scale is $d/L=1/60$, yieldingThe finite-distance scaling can also be checked without treating an intensity pattern as a collection of geometrical point objects: the slit-to-screen Fresnel integral after a thin lens has quadratic coefficient $1/d-1/f=1/L$ and linear term $-y/d$. It therefore reproduces the old diffraction pattern at transverse coordinate $Y=Ly/d$, up to an overall amplitude and phase. If the old far-field plane is treated as infinitely distant, the equally useful leading approximation is $f=d$, $R=6.0\mathrm{cm}$; the finite-distance correction is only $1.7\%$.

The colour discussion below assumes a smooth white incident spectrum and neglects lens dispersion when comparing the ideal diffraction patterns.

In series the same current passes through both resistors, while their voltage drops add: $V=I(R_1+R_2)$. In parallel the same voltage drives currents $V/R_1$ and $V/R_2$, which add. Thus $R_s=R_1+R_2$ and $R_p^{-1}=R_1^{-1}+R_2^{-1}$.

Use phasors proportional to $e^{i\omega t}$. Then $Z_L=i\omega L$ and $Z_C=1/(i\omega C)$. The printed diagram places C in parallel with the series RL branch, so the input admittance isIts nonzero zero-phase frequency exists for $R^2<L/C$:At zero frequency, $|Y|=1/R$ and $I_0=V_0/R$. At ${\omega }_1$, $R^2+{\omega }_1^2{L}^2=L/C$, givingFor $0<\omega <{\omega }_1$, $\mathrm{Im}Y<0$, so current lags voltage. At very high frequency, $Y\sim i\omega C$, so current leads by $\pi /2$. Zero phase need not coincide exactly with the minimum current magnitude in this lossy network.

The three measured limits determine the components in order:For the changed resistance, the quoted $0.6\mathrm{kHz}$ is an ordinary frequency, so ${\omega }^{\prime }=2\pi (600{\mathrm{s}}^{-1})$. Keeping L and C fixed gives

Measure $x$ from the static equilibrium position, so the constant gravitational force has already cancelled the equilibrium spring force. With stationary oil, Newton's second law gives $m\ddot{x}=-b\dot{x}-kx$. The characteristic exponents are $-\gamma \pm i{\omega }_1$, whereFor the stipulated phase of the damped cosine,At $t_n=2\pi n/{\omega }_1$, the velocity is not zero: $\dot{x}=-\gamma A{e}^{-\gamma t_n}$. ThusExpanding $e^{-z}$ at small $z=4\pi \gamma /{\omega }_1$ gives the requested light-damping loss $z$. The exact instantaneous energy also satisfies $dE/dt=-b{\dot{x}}^2\le 0$.

From the static extension, $k=mg/(0.20\mathrm{m})=44.145\mathrm{N}{\mathrm{m}}^{-1}$; the instantaneous drag measurement gives $b=0.10\mathrm{N}\mathrm{s}{\mathrm{m}}^{-1}$. Consequently $\gamma =0.05556{\mathrm{s}}^{-1}$ and ${\omega }_1=7.003{\mathrm{s}}^{-1}$. Two successive downward equilibrium crossings are one damped period apart, so the same energy ratio applies. At the first crossing $E=m{v}^2/2=1.125\times 10^{-3}\mathrm{J}$. ThereforeThe speed has decreased by $2.43\times 10^{-3}\mathrm{m}{\mathrm{s}}^{-1}=0.243\mathrm{cm}{\mathrm{s}}^{-1}$. These values use the exact decay factor, not the first-order approximation.

For a fixed rotation axis with constant moment of inertia, the axial torque obeys $\tau =I\ddot{\theta }$, with $\tau =\sum (\mathbf{r}\times \mathbf{F})\cdot {\hat{\mathbf{e}}}_{\mathrm{axis}}$. For a point mass at radius $l$, the tangential equation is $F_t=ml\ddot{\theta }$; multiplying by $l$ gives $l{F}_t=m{l}^2\ddot{\theta }$, as required. Radial forces have zero moment about the axis.

Take increasing $\theta$ in the direction of the clapper's initial fall and let the contact force $N$ oppose that tangential motion. Initial equilibrium of the clapper givesWhile contact persists the bell and clapper rotate together. Their respective offsets from the upward vertical are $\theta -\beta$ and $\theta$. Summing external gravitational moments eliminates their internal contact force:The clapper alone satisfies $m{l}^2\ddot{\theta }=mgl\sin \theta -Nl$. Hence for $m\ll M$,A nonadhesive contact cannot exert tension. The first zero of this expression on the released trajectory is the loss-of-contact angle. Write $q=lh/k^2=0.9375$. ThenThe second-quadrant root is reached from $\theta =\beta =30^{\circ }$; selecting the principal negative arctangent would choose the wrong branch. Indeed the exact contact force before taking $m/M\to 0$ is $mMg[k^2\sin \theta -lh\sin (\theta -\beta )]/(M{k}^2+m{l}^2)$, so its zero is unchanged.

Take inertial axes parallel, with frame $S^{\prime }$ moving at $+v\hat{\mathbf{x}}$ relative to $S$ and origins coincident at zero time. The Lorentz factor is ${\gamma }_v=(1-v^2/c^2{)}^{-1/2}$; $\Delta x,\Delta y,\Delta t$ are separations of the same pair of events in $S$, and primed quantities are their separations in $S^{\prime }$. The transverse coordinates are unchanged. For two events on an inertial clock's worldline,The proper time $\Delta \tau$ is that clock's elapsed time; uniform speed $w$ gives $\Delta t={\gamma }_w\Delta \tau$. Coordinate intervals between spatially separated events must instead be Lorentz-transformed with their spatial separation included.

For the round trip in the station frame, the rendezvous distance is $u{t}_{\mathrm{out}}$, and the return has the same speed and distance. Thus $t_{\mathrm{round}}=2t_{\mathrm{out}}=10a/(\sqrt{7}c)$. Its departure and arrival both occur at the station, so $\Delta x=0$ for these endpoints:This uses the return speed as measured by the station. The elapsed time aboard A, apart from an ideal instantaneous turn, would be $\Delta t_{S,\mathrm{round}}/{\gamma }_u$.

For the crash, A's station-frame velocity components are $(3c/5,\sqrt{7}c/5)$, while B's are $(3c/5,0)$. Here ${\gamma }_u=5/3$ and ${\gamma }_v=5/4$. Four-momentum conservation, with no separately escaping radiation, gives the wreckage totals in the station frame:In B's original inertial frame the same totals areThe invariant wreckage mass is $m\sqrt{14/3}$, exceeding $2m$ because internal energy remains in the wreckage. Its momentum cannot be specified without naming a frame; both natural frames have been given.

In the centre-of-momentum frame the two equal masses have opposite velocities. Conservation of kinetic energy in an elastic collision fixes their common speed, so scattering changes their directions only. For incoming laboratory velocities $\mathbf{v}$ and $0$, the frame velocity is $\mathbf{V}=\mathbf{v}/2$ and the incoming relative velocities are $\pm \mathbf{v}/2$. If the outgoing relative velocities are $\pm \mathbf{w}$, then $|\mathbf{w}|=v/2$ and the laboratory velocities are $\mathbf{V}\pm \mathbf{w}$. Their dot product is zero: nonzero outgoing laboratory velocities are perpendicular.

For this smooth equal-disc oblique collision, let $\mathbf{n}$ point from A to B at contact. The impulse acts only along $\mathbf{n}$; the tangential velocity of A is unchanged. With $\mathbf{v}\cdot \mathbf{n}=v\cos \theta$, equal masses exchange their normal velocity components:ThereforeNo rotation about a disc centre is generated by these central impulses. At exactly head-on impact A stops, so its outgoing direction is undefined; the angle formula then denotes a limiting direction.

For this adhesive collision of two equal discs, use the parallel axis theorem about the midpoint of the touching centres:Its magnitude is $mav\sin \theta$. Linear and angular momentum conservation giveThe rotation has the sign of ${\mathbf{L}}_{\mathrm{CM}}$, clockwise for the printed geometry. The energy loss is $m{v}^2(1/4-{\sin }^2\theta /6)>0$. After impact the midpoint translates uniformly and the rigid pair rotates uniformly about it: there are no external horizontal forces or axial torques.

Rough unglued edges can transmit a tangential impulse and generate angular momentum about the individual disc centres. The centre-of-mass velocity and total angular momentum still have the values above, but individual translational speeds, angular velocities and energy loss require a friction/contact law and normal restitution. Roughness alone does not imply permanent sticking or the adhesive result.