For steady electric currents in vacuum, Ampère's circuital law isChoose an oriented closed contour and a spanning surface; the positive enclosed electric current crosses that surface according to the right-hand rule for the contour. First use the source symmetry to establish the direction of $\mathbf{B}$ and which positions have the same magnitude. Then choose the contour so its useful segments are parallel to $\mathbf{B}$ with constant magnitude, and the remaining segments are perpendicular to $\mathbf{B}$ or have vanishing field contribution. The line integral reduces to the unknown magnitude times a known length or sum of lengths.

For example, cylindrical symmetry around an infinite wire allows a circular contour, giving $B(2\pi r)={\mu }_{0}I$. A uniform current sheet instead calls for a rectangle straddling it. The law is always valid in this magnetostatic setting, but by itself does not determine the local magnetic field from an arbitrary enclosed current without these extra symmetry arguments. Here ${\mu }_0$ is the vacuum permeability.

For a downward step potential, $V_0=-|V_0|$, the region on the right is classically allowed at every positive incident energy. The same matching calculation applies withAs $E\to 0^{+}$, $k_1\to 0$ and $k_2\to k_{\ast }=\sqrt{2m|V_0|}/\hslash >0$. Henceand the flux coefficients satisfyThis quantum reflection at a downward step occurs even though a classical particle would simply accelerate into the lower-potential region. The abrupt change in wavenumber causes strong reflected-wave interference.

Both this limit and part (d) give total reflection, but their probability amplitudes and spatial densities differ. At the upward-step threshold, $B/A\to +1$ and $C/A\to 2$: an antinode meets the origin and the right-hand density stays finite while its propagation velocity vanishes. Here $B/A\to -1$ and $C/A\to 0$: the incident and reflected waves nearly cancel at the origin, and the transmitted density vanishes while the right-hand propagation velocity remains finite. In both descriptions the incident probability amplitude is held fixed for comparison.

The coefficients above are limiting flux ratios for $E>0$. At exactly $E=0$ the incident probability current vanishes, so directly dividing zero-energy fluxes would not define a scattering probability.

Use adjacent segments $[-\ell ,0]$ and $[0,\ell ]$, with $\ell =2\pi /k_1$. Integrate the threshold probability density from part (d). The left segment contains two complete density periods, so its average ${\cos }^2$ is $1/2$:The right-hand density is constant:The common stationary-state normalization cancels, yieldingThe larger local probability weight on the right is compatible with zero transmitted probability current at threshold: density and flux measure different quantities.

Take the threshold from the above-step side used in the preceding parts, $E\to V_0^{+}$. Then $k_2\to 0$ while $k_1\to \sqrt{2m{V}_0}/\hslash$ remains finite. The reflected probability amplitude approaches $B=A$, and the transmitted amplitude approaches $C=2A$. At each fixed position, the potential-step threshold wavefunction is thereforeIts probability density is

The left-hand standing wave has zero-density nodes at $k_{1}x=-\pi /2,-3\pi /2,\dots$ and maxima of height $4|A{|}^2$. Its density period is $\pi /k_1$, half the incident de Broglie wavelength. At the origin the density is a maximum, continuous with the constant value on the right; its first derivative is also zero on either side there.

The nonzero right-hand density does not imply transmitted flux: $j_{\mathrm{tr}}=\hslash k_2|C{|}^2/m\to 0$, so $R\to 1$ and $T\to 0$. The exact threshold solution with no incoming component from the right is the limiting constant solution of ${\psi }^{\prime \prime }=0$ in the raised region. These stationary scattering states are not normalized over the whole real line; finite-interval density ratios are nevertheless well defined.

For $E\gg V_0$, the two wavenumbers nearly coincide:Consequently $B/A\simeq V_0/(4E)\to 0$ and $C/A\to 1$. To leading order the complete wavefunction isThe small fractional change in kinetic energy scarcely changes the propagation wavenumber, so reflection vanishes and the particle behaves approximately as a free right-moving plane wave. A more accurate phase at large positive $x$ retains $k_{2}x$ on the transmitted side; even a small wavenumber difference accumulates over a sufficiently long distance.

The probability current of the incident wave is $j_{\mathrm{in}}=\hslash k_1|A{|}^2/m$, and that of the reflected wave is $j_{\mathrm{ref}}=-\hslash k_1|B{|}^2/m$. The quantum reflection probability is the positive ratio of reflected to incident flux magnitudes:For comparison, the transmitted probability current is $j_{\mathrm{tr}}=\hslash k_2|C{|}^2/m$, so the transmission probability isThese satisfy $R+T=1$, as required by conservation of probability current for a real, nonabsorbing potential energy. In particular, $|C/A{|}^2$ alone is not the transmitted flux fraction when $k_2\ne k_1$.

Use a stationary wavefunction $\Psi (x,t)=\psi (x)e^{-iEt/\hslash }$ solving the time-independent Schrödinger equation. For energy above the positive step potential, the positive wavenumbers areIncidence only from the left permits a right-moving incident plane wave and a left-moving reflected plane wave for $x\le 0$, but only a right-moving transmitted plane wave for $x>0$:The omitted right-incident component would describe a different scattering experiment. Wavefunction matching at a finite potential step requires continuity of both $\psi$ and $d\psi /dx$ for a finite jump and the same mass on either side:Eliminate $B=C-A$ from the second condition:The first ratio is a transmission amplitude; a transmitted probability current also depends on its different propagation velocity.

For a light test particle of mass $m$ outside a spherical source of mass $M$, Newton's law of universal gravitation gives $\mathbf{F}=-GMm\hat{\mathbf{r}}/r^2$. It is a central force, so its torque about the source vanishes and angular momentum $\mathbf{L}=m\mathbf{r}\times \mathbf{v}$ is conserved. The motion therefore lies in a fixed plane when $L\ne 0$. Its gravitational potential energy is $-GMm/r$, and conservation of mechanical energy givesHere $ϵ$ is specific orbital energy and $h$ is specific angular momentum. A Kepler orbit has polar equationThe angle is measured from periapsis. The conic section is an ellipse for $ϵ<0$ and $0\le e<1$, a parabola for $ϵ=0$ and $e=1$, and a hyperbola for $ϵ>0$ and $e>1$. The circle is the $e=0$ elliptical case. The special case $h=0$ is radial motion and must be handled as a degenerate limit. Bound ellipses have the source at one focus, which is Kepler's first law.

For a bound orbit of semi-major axis $a$, $ϵ=-GM/(2a)$. Combining this with the energy equation gives the vis-viva equation:The particle therefore moves faster near periapsis and slower near apoapsis. Conservation of angular momentum also gives constant areal velocity, $dA/dt=h/2$, expressing Kepler's second law. Using the ellipse's area and $h^2=GMa(1-e^2)$ gives the orbital periodwhich is Kepler's third law.

A circular orbit follows from centripetal acceleration $v_c^2/r=GM/r^2$, giving $v_c=\sqrt{GM/r}$ and $T=2\pi \sqrt{r^3/(GM)}$. The minimum escape velocity from radius $r$ follows from the marginally unbound condition $ϵ=0$:These ideal orbits neglect drag force, other gravitating bodies and changes in the central source. For two comparable masses, the relative two-body problem uses $G(M+m)$ instead of $GM$. An orbiting object is continuously in free fall; the inward gravitational force changes its direction even when its speed is constant. These inverse-square results assume the physical variation of gravity with radius, rather than extending a near-surface constant-$g$ approximation to an entire orbit.

Faraday's law of induction relates an induced electromotive force to the rate of change of magnetic flux. Choose a positive orientation around a loop and its associated surface normal by the right-hand rule. For a stationary conducting loop,Unlike an electrostatic field, the induced electric field generally has nonzero curl; the differential form is $\nabla \times \mathbf{E}=-\partial \mathbf{B}/\partial t$. For a moving circuit, magnetic flux transport for a moving circuit includes the magnetic Lorentz force on the charges:where $\mathbf{u}$ is the local wire velocity. Thus changing the field or moving a conductor can produce induction.

The minus sign is Lenz's law: an induced electric current, when a closed conducting path permits it, produces a magnetic field opposing the change of linked flux. It need not oppose the original field itself. For example, increasing flux through a loop produces a current tending to reduce that increase; decreasing the same flux reverses the current. The opposition is consistent with conservation of energy: external mechanical or electrical work is needed to maintain a change against its inductive reaction.

For a coil of $N$ turns, define the magnetic flux linkage $\Lambda ={\sum }_{j=1}^N{\Phi }_j$, reducing to $N{\Phi }_B$ when every turn links the same flux. For fixed geometry and a linear magnetic response, self-inductance $L$ is defined by $\Lambda =LI$, and the coil's induced electromotive force isThe corresponding terminal voltage drop, with the passive sign convention, is $V_L=LdI/dt$. It opposes changes in electric current; it does not require the current itself to vanish.

Building the current from zero requires power $P=I{V}_L=LIdI/dt$. Integrating gives the energy stored by an ideal inductor:This energy is stored in the magnetic field and can be returned when the current falls. In a series RL circuit driven by a constant voltage $V_0$, Kirchhoff's voltage law gives $L\dot{I}+RI=V_0$. Starting with zero current, $I=(V_0/R)(1-e^{-tR/L})$, with time constant $L/R$. This makes the resistance to rapid current changes directly observable. Two coils also have mutual inductance: changing primary current induces secondary electromotive force $-Md{I}_1/dt$ with a polarity determined by the chosen winding orientations, the principle behind a transformer.

A conservative vector field has path-independent work: the integral ${\int }_A^B\mathbf{F}\cdot d\mathbf{r}$ depends only on its endpoints. Equivalently, the line integral around every closed path in its domain vanishes. One can then define a single-valued potential energy $U$ byThe arbitrary additive constant in $U$ has no effect on the force. A useful zero is often chosen at infinity for a localized attractive interaction.

A differentiable conservative force has $\nabla \times \mathbf{F}=0$ because the curl of a gradient vanishes. Conversely, a smooth curl-free field on a simply connected domain admits a global scalar potential. The domain condition matters: on the punctured plane, $\mathbf{F}=(C/r)\hat{\mathbf{\theta }}$ has zero curl away from the missing origin but closed-loop work $2\pi C$ around it. It is not globally conservative there. The existence of a local potential does not establish a single-valued potential throughout a domain with holes.

The work-energy theorem gives $\Delta T=W$ for kinetic energy $T$. If all forces are conservative and $U$ has no explicit time dependence, thenThis is conservation of mechanical energy. In one dimension, an equilibrium occurs at $dU/dx=0$; a local minimum with $U^{\prime \prime }>0$ is stable, while a maximum with $U^{\prime \prime }<0$ is unstable. Expanding about a minimum gives $U\simeq U_0+U^{\prime \prime }(x_0)(x-x_0{)}^2/2$, producing simple harmonic motion with ${\omega }^2=U^{\prime \prime }(x_0)/m$.

Examples include the spring force $F_x=-kx$ with $U=k{x}^2/2$, the near-surface gravitational force $-mg\hat{\mathbf{z}}$ with $U=mgz$, and the central Newtonian gravitational field with $U=-GMm/r$. An electrostatic force on a fixed electric charge $q$ has $U=q\Phi$, where $\Phi$ is the electric potential.

Sliding friction provides a contrasting physical test: traversing a closed path still dissipates mechanical energy, since the frictional force opposes the local motion on both outward and return legs. Its work depends on path length and cannot be represented by a position-only $U$. Total energy can remain conserved when the resulting internal energy is included. Also, even if $\mathbf{F}=-\nabla U(\mathbf{r},t)$ at each instant, explicit time dependence gives $d(T+U)/dt=\partial U/\partial t$; a static conservative-field argument alone does not guarantee constant mechanical energy in that situation.

An electric dipole is a localized pair of opposite electric charges, or more generally a neutral electric charge distribution with nonzero first electric moment. For charges $-q$ and $+q$ separated by displacement $\mathbf{d}$ directed from negative to positive, the electric dipole moment isA neutral molecule can have a permanent electric dipole moment when its positive and negative charge centres do not coincide. An applied electric field can also produce an induced electric dipole moment by distorting an initially nonpolar electric charge distribution.

At distance $r\gg |\mathbf{d}|$ in vacuum, expand the difference of the two point-charge potentials to first order in the separation. The leading electric potential isThe monopole term cancels because the net electric charge is zero. Taking $\mathbf{E}=-\nabla \Phi$ gives the electric dipole fieldOn the dipole axis its magnitude is $2p/(4\pi {\epsilon }_0{r}^3)$ and its direction is along $\mathbf{p}$; in the equatorial plane the magnitude is $p/(4\pi {\epsilon }_0{r}^3)$ and the field points opposite to $\mathbf{p}$. The $r^{-3}$ decay is faster than the $r^{-2}$ field of an isolated point charge.

In a uniform external electric field, the opposite forces on the two charges cancel in their resultant but form a couple. Its torque isFor a fixed permanent electric dipole moment, the interaction potential energy isDifferentiating with respect to orientation recovers the torque. Parallel alignment minimizes $U$ and is stable against a small rotation; antiparallel alignment maximizes $U$ and is unstable. Rotating a fixed dipole slowly from parallel to antiparallel requires external work $2pE$.

A nonuniform electrostatic electric field can also exert a net force. Expanding the two charge forces gives $\mathbf{F}\simeq (\mathbf{p}\cdot \nabla )\mathbf{E}$, which equals $\nabla (\mathbf{p}\cdot \mathbf{E})$ for a fixed dipole in a curl-free electrostatic field. Thus a suitably aligned dipole is attracted toward stronger field. An induced electric dipole moment requires care in computing energy: when $\mathbf{p}=\alpha \mathbf{E}$ for constant isotropic electric polarizability, building the dipole costs energy and the effective interaction is $U=-\alpha E^2/2$, rather than substituting directly into the fixed-dipole formula. These field, torque and energy effects explain molecular orientation and electric polarization.

In SI units, the microscopic Maxwell's equations for total charge density $\rho$ and current density $\mathbf{J}$ areThe first is Gauss's law: electric charge sources electric flux. The second is Gauss's law for magnetism, expressing the absence of magnetic monopole sources in this theory. Faraday's law relates changing magnetic flux to circulating electric field. The Ampere-Maxwell law relates magnetic circulation to both material current and changing electric flux.

For fixed loops and surfaces, the integral versions areSurface and loop orientations are related by the right-hand rule. These forms follow from the divergence theorem and Stokes' theorem. In material media, equivalent macroscopic equations can separate free and bound sources using $\mathbf{D}$ and $\mathbf{H}$ together with constitutive relations; the displayed forms use total sources.

The displacement-current term is essential for charge conservation. Taking the divergence of the last differential equation and using Gauss's law gives ${\partial }_t\rho +\nabla \cdot \mathbf{J}=0$. It also makes magnetic circulation consistent for surfaces spanning the same loop through either a charging capacitor's wire or its dielectric gap.

In source-free vacuum the curl equations combine to give ${\nabla }^2\mathbf{E}={\mu }_0{ϵ}_0{\partial }_t^2\mathbf{E}$, with the same equation for $\mathbf{B}$. ThusA plane wave is transverse, with $\mathbf{E}$, $\mathbf{B}$ and propagation direction mutually perpendicular and $E=cB$. The fields transport energy with density $u={ϵ}_0{E}^2/2+B^2/(2{\mu }_0)$ and Poynting vector $\mathbf{S}=\mathbf{E}\times \mathbf{B}/{\mu }_0$. Together with the Lorentz force law, the equations connect charges, currents, forces, induction and electromagnetic radiation.

For steady currents in vacuum, the Biot-Savart law gives the magnetic field by summing oriented current elements. With $\mathbf{R}=\mathbf{r}-{\mathbf{r}}^{\prime }$ from the source to the observation point,For a thin wire it becomes $\mathbf{B}=({\mu }_{0}I/4\pi )\int d{\mathbf{l}}^{\prime }\times \hat{\mathbf{R}}/R^2$. The cross product fixes the direction by the right-hand rule. Fields from different steady circuits add by superposition; the entire current path, including return wires where relevant, belongs in the calculation.

An infinite straight wire gives circular field lines of magnitude $\boxed{B={\mu }_{0}I/(2\pi s)}$ at perpendicular distance $s$. This follows by integrating the transverse component of each element along the wire. A circular loop of radius $a$ has cancelling transverse components on its axis, leavingMultiple coincident turns multiply the field by their number. Far from the loop this is the field of a magnetic dipole of moment $\mathbf{\mu }=I\pi a^2\hat{\mathbf{n}}$. Integrating over many loops gives the approximately uniform interior field $B\simeq {\mu }_{0}nI$ of a long solenoid, with $n$ turns per unit length. Finite coils have fringing fields, and appropriately spaced coil pairs can produce a more uniform measurement region.

The law is useful for predicting fields of wires and coils where direct symmetry is insufficient, designing electromagnets and estimating stray fields. It is a magnetostatic law: steady charge flow satisfies charge continuity. An instantaneous application to rapidly changing currents neglects propagation and radiation; such problems require the time-dependent Maxwell's equations and retarded fields. A quasistatic approximation can be valid when circuit dimensions are small compared with the light-travel distance during the variation timescale.

A position-dependent force is a conservative force when its work depends only on endpoints. Equivalently, its work around any closed allowed path vanishes, and one can define a potential energy $U$ satisfyingThe potential is defined up to an additive constant. A vanishing curl, $\nabla \times \mathbf{F}=0$, is a local test; it guarantees a global single-valued potential on a simply connected region with suitable regularity. Holes or excluded singularities require checking the global path condition.

The work-energy theorem gives $dK/dt=\mathbf{F}\cdot \mathbf{v}$. For a static potential this implies $d(K+U)/dt=0$. If $U$ depends explicitly on time, then instead $d(K+U)/dt=\partial U/\partial t$ under that force alone. Friction and other nonconservative forces add their own power and can convert mechanical energy into heat.

Examples include the gravitational potential energy $-GMm/r$ outside a spherical mass and the electrostatic energy $U=qV$, where $\mathbf{E}=-\nabla V$. For a point source charge $Q$, $U=qQ/(4\pi {ϵ}_{0}r)$, giving attraction or repulsion through the sign of $qQ$. A spring has $U=k{x}^2/2$ and force $-kx$. An equilibrium has $\nabla U=0$; a strict local minimum is stable against small displacements when the quadratic curvature is positive. Expanding about that minimum explains the harmonic restoring force used in many small-oscillation models.

Magnetic force on a moving point charge does no work, but it is velocity-dependent and is not generally a position-only force $-\nabla U$. Likewise a time-varying magnetic flux induces a circulating electric field with $\oint \mathbf{E}\cdot d\mathbf{l}=-d{\Phi }_B/dt$, so that electric field cannot globally be represented solely by an electrostatic scalar potential. These distinctions prevent treating every field force as conservative simply because an energy description exists for the complete system.

A charge $q$ moving nonrelativistically in prescribed fields obeys the Lorentz force lawThe magnetic part is perpendicular to velocity, so $d(m{v}^2/2)/dt=q\mathbf{E}\cdot \mathbf{v}$: the electric field can change kinetic energy, while the magnetic field alone changes direction without doing work.

In a uniform electric field with no magnetic field, acceleration is constant: $\mathbf{v}={\mathbf{v}}_0+(q\mathbf{E}/m)t$ and $\mathbf{r}={\mathbf{r}}_0+{\mathbf{v}}_{0}t+q\mathbf{E}{t}^2/(2m)$. The trajectory is a parabola when there is initial motion perpendicular to the field, and a line when motion is purely along it. Reversing the charge reverses the electric acceleration.

In a uniform magnetic field with no electric field, resolve velocity into parallel and perpendicular components. The parallel component is constant; the perpendicular component rotates at the cyclotron frequencyThe resulting path is a helix, or a circle if $v_{\parallel }=0$, with pitch $2\pi v_{\parallel }/{\omega }_c$. Charge sign determines its sense of rotation. These relations underlie magnetic bending and mass/charge analysis. At relativistic energies, use momentum $\mathbf{p}=\gamma m\mathbf{v}$; at fixed energy the cyclotron frequency becomes $|q|B/(\gamma m)$.

For uniform crossed fields $\mathbf{E}\perp \mathbf{B}$, the velocity ${\mathbf{v}}_D=\mathbf{E}\times \mathbf{B}/B^2$ satisfies $\mathbf{E}+{\mathbf{v}}_D\times \mathbf{B}=0$. The orbit is a gyromotion about a guiding center drifting withindependent of mass and charge sign. A velocity selector transmits particles with the appropriate direction and speed $E/B$ without deflection. If the electric field has a component along the magnetic field, that component accelerates the parallel motion instead. Nonuniform fields can add other drifts; the simple helix and uniform crossed-field expressions require the stated uniform-field assumptions.

The examination asks for three notes. All four options are worked below so that each topic has a complete study answer.