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For a steady current density $J$, the magnetic field is

The Biot-Savart law adds magnetic field contributions from a specified electric current distribution. The vectors ${\mathbf{r}}^{\prime }$ and $\mathbf{r}$ locate source and observer; $\mathbf{r}-{\mathbf{r}}^{\prime }$ points toward the observer. Here ${\mu }_0$ is vacuum permeability. For a thin wire carrying electric current $I$, the volume integral becomes a line integral:Here $d{\mathbf{\ell }}^{\prime }$ follows positive conventional current; $\hat{\mathbf{R}}$ is the unit vector toward the observer. With fixed path orientation, reversing the electric current replaces $I$ by $-I$. The cross product and right-hand rule fix the direction. Reversing the source-to-observer displacement would give the wrong sign.^[ref]

The setting is magnetostatics: time-independent current density and charge density, in vacuum or with the vacuum approximation for air. Electric charge conservation then gives $\nabla \cdot \mathbf{J}=0$. A finite segment is part of a complete electrical circuit; its endpoints cannot continuously accumulate electric charge in a steady state. The instantaneous integral above does not describe switching transients or electromagnetic radiation. In magnetic matter, specify the total current density, including magnetization current, rather than automatically replacing ${\mu }_0$ by an arbitrary magnetic permeability.^[ref]

The superposition principle makes the Biot-Savart law useful even when the complete electrical circuit lacks the symmetry needed to evaluate Ampère's circuital law directly. Integrate a straight piece, a curved piece, and the return path separately, then add their magnetic fields as vectors. The beauty is that a local oriented current element produces both the circulating magnetic field of a finite straight wire and the reinforcing on-axis magnetic field of a circular current loop.

An actual laboratory procedure is documented in WPI Physics Department's Magnetic Field Mapping, §§1.2–1.2.3 and the single-coil checklist on p. 9. Preparation locates the active area of a Hall effect sensor using a small magnet. A field meter reads the component normal to that area. The wire apparatus is a wall-mounted square loop selectable between one and five conductors. The coil connects to a power supply and ammeter; its diameter and turn count are recorded. Align the sensor and scan distance. The control is a current-off background reading at each position, subtracted from the current-on reading. The inference compares corrected magnetic field with position and the predicted geometry dependence. Active-area displacement, sensor tilt, background variation, and finite coil width limit that comparison. These are instructions, not reported measurements.^[ref]

A further proposed control is to reverse $I$ at fixed sensor position and orientation. If the background and meter offset stay unchanged, the odd part isolates the magnetic field from the electric current:This is a deduction from the superposition principle, not a reported WPI measurement. The measured component, sensor position, and complete return path must still match the model.

An integral over the boundary of a region, such as a line integral around a planar region or a surface integral over the boundary of a three-dimensional region. Green's identities convert certain volume integrals into boundary integrals.

Spherical symmetry makes moments over a ball isotropic tensors. Odd moments vanish under $x\mapsto -x$, while the second and fourth moments areFor the second moment, a tensor contraction gives $3\alpha =\int r^{2}dV=4\pi a^5/5$. For the fourth, complete index symmetry makes the three coefficients equal and a double tensor contraction gives $15\beta =\int r^{4}dV=4\pi a^7/7$. These identities turn polynomial volume integrals into scalar radial integrals and are useful for inertia and rotational averages.

A current element is the oriented infinitesimal wire contribution $Id{\mathbf{\ell }}^{\prime }$ inside the Biot-Savart law. For uniform axial current density across wire cross-sectional area $A$, $I=JA$ and $d^3{r}^{\prime }=Ad{\ell }^{\prime }$ imply $\mathbf{J}{d}^3{r}^{\prime }=Id{\mathbf{\ell }}^{\prime }$. This reduces a volume integral to a line integral in the thin-wire approximation. It is an integration contribution, not an isolated steady source. Conventional current runs opposite the drift of negatively charged electrons.

A current loop is a closed path carrying electric current. For a thin wire following a closed curve $C$, the volume integral defining its magnetic dipole moment becomes a line integral:The vector area $\mathbf{A}$ is $A\hat{\mathbf{n}}$ for a planar current loop. For a nonplanar current loop, use the vector area, rather than the scalar area of an arbitrarily chosen spanning surface. Reversing the electric current reverses $\mathbf{m}$ without changing the geometry. ^[ref].

For $N$ closely spaced identical turns carrying the same electric current, the magnetic dipole moments add: $\mathbf{m}=NIA\hat{\mathbf{n}}$. Each turn contributes its own vector area; unequal areas or differently oriented turns require a vector sum. As a calculated example, twelve circular turns of radius $R=3.0\mathrm{cm}$ carrying $I=0.40\mathrm{A}$ give $m=NI\pi R^2=1.36\times 10^{-2}\mathrm{A}{\mathrm{m}}^2$. This is an idealized example, not an experimental measurement.

For $x=r\cos \varphi$, $y=r\sin \varphi$ and unchanged $z$, the Jacobian determinant of Cartesian coordinates with respect to cylindrical coordinates is $r$. Hence a volume integral acquires the displayed factor for $r\ge 0$. The inverse determinant is $1/r$ where $r>0$; the axis is a coordinate degeneracy.

For the outward spherical patch $r=R$, $y\le 0$, $z\le 0$, and the polynomial vector field $F=(y{x}^2,x{y}^2-3z^4,x^3+y^2)$, close the quarter-ball with its two planar half-discs. The divergence is $4xy$, whose volume integral vanishes by symmetry in $x$. The caps have fluxes $-3\pi R^6/16$ and $\pi R^4/8$, obtained from half-disc fourth and second moments. The divergence theorem gives the displayed spherical flux. All pieces use normals pointing out of the same closed volume; the caps are auxiliary and must be subtracted.

The electric field does work on matter at local power per volumeHere $\mathbf{J}$ is current density and the dot product includes the relative direction. Positive $\mathbf{J}\cdot \mathbf{E}$ means energy passes from the field to matter; negative means matter supplies energy to the field. In general this is a work rate, not automatically an irreversible heating rate.^[ref]

For a stationary isotropic Ohmic conductor with real scalar electrical conductivity $\sigma >0$, local Ohm's law gives $\mathbf{J}=\sigma \mathbf{E}$. Thus the local Joule heating rate isThe electrical resistivity ${\rho }_{\mathrm{el}}$ has units $\Omega \mathrm{m}$, and electrical conductivity has units $(\Omega \mathrm{m}{)}^{-1}$. These formulas use the material's rest frame and dissipative conduction current density. Moving media, complex conductivity, and additional storage or transport processes require their appropriate energy balance rather than this scalar substitution.

For uniform current in a straight homogeneous wire of length $L$, wire cross-sectional area $A$ and constant electrical conductivity, $E=V/L$, $J=I/A$ and $R=L/(\sigma A)$. The volume integral recovers terminal power:As a numerical check, the wire already used in electrical resistance has $L=1\mathrm{m}$, $A=10^{-6}{\mathrm{m}}^2$, $\sigma =5\times 10^5(\Omega \mathrm{m}{)}^{-1}$ and $V=5\mathrm{V}$. It gives $J=2.5\times 10^{6}A/{\mathrm{m}}^2$, $p_{\mathrm{vol}}=1.25\times 10^{7}W/{\mathrm{m}}^3$ and $P=12.5\mathrm{W}$, agreeing with $I=2.5\mathrm{A}$ and $R=2\Omega$. Concentrating the same electric current in a smaller wire cross-sectional area therefore raises the local heating rate if electrical resistivity stays fixed.

For a solid spherical cap $r^2+z^2\le R^2$, $z\ge c$, where $0\le c\le R$ and the mass density is $\rho =kz$ with $k\ge 0$, a volume integral in cylindrical coordinates givesHere the height is measured from the centre of the sphere, rather than from the bottom of the cap. Specifying that origin matters for a nonuniform mass density.

The gravity-capillary drop with a meridional curvature approximation uses the stipulated $\kappa =-h^{\prime \prime }$ for this calculation. At equilibrium the thin-film equation givesRegularity at $r=0$ forces $C=0$. In the wetted interior $h>0$, the equilibrium pressure gradient therefore vanishes, givingThis is the capillary length. Integrating the constant-coefficient ordinary differential equation givesThe central boundary condition $h^{\prime }(0)=0$ sets $C_2=0$, and contact at $r=R$ gives $C_0=-C_1\cosh (R/l)$. Write $C_1=-\alpha l$. The prescribed contact angle fixesHere $\alpha$ is the dimensionless coefficient of this drop solution, unrelated to the hinge angle in question 1.

The axisymmetric volume integral is $V=2\pi {\int }_0^{R}rh(r)dr$. Substituting $\xi =r/l$ givesFor numerical evaluation, with $s=R/l$, the integral isTogether with $\alpha =\tan {\theta }_c/\sinh s$, this determines $R$ for fixed $V$ and ${\theta }_c$.

The Bond number $R^2/l^2=\rho g{R}^2/\gamma$ distinguishes the dominant balances. For $R\ll l$, surface tension dominates gravity and the surface is nearly a small-slope constant-curvature cap:For $R\gg l$, gravity flattens the interior, while surface tension bends the edge over the capillary length. The stipulated model givesup to relative corrections of order $l/R$. The contact angle fixes the edge slope and therefore the height scale; conservation of mass then fixes the lateral extent. Consistency of lubrication theory requires small slopes, in particular $\tan {\theta }_c\ll 1$.

The stated approximation differs from axisymmetric free-surface curvature. For an axisymmetric graph with the same sign convention, the sum of principal curvatures isThe omitted $h^{\prime }/r$ term is generally of the same order as $h^{\prime \prime }$. For example, $h=H(1-r^2/R^2)$ gives stipulated $\kappa =2H/R^2$ but axisymmetric small-slope $\kappa =4H/R^2$. Thus the required hyperbolic-cosine profile solves the explicitly prescribed reduced model, not the full axisymmetric capillary equilibrium. Keeping both small-slope terms instead giveswhere $I_0,I_1$ are modified Bessel functions. The qualitative capillary length and Bond number balances remain valid.

Take the normal vector to point outward from the box. The divergence theorem changes the closed-surface flux into a volume integral. The divergence isThe first term integrates to zero in $x$, since ${\int }_{-1}^2\pi \cos (\pi x)dx=\sin 2\pi -\sin (-\pi )=0$. ThereforeHenceA direct face check gives zero from the two $x$ faces, $45/2$ from the pair of $y$ faces, and $45$ from the pair of $z$ faces. This also checks the outward-orientation signs.

A volume mean is the volume integral of the temperature divided by the volume. On this cube,Each one-dimensional integral is two, givingThe number is in the temperature units implicit in the given field law and coordinate units.

Use the outward normal on the spherical patch. It is the quarter of the spherical surface lying below both the horizontal plane and the plane $y=0$, with both signs of $x$ included. Close it by the two half-discs $D_y$ in $y=0,z\le 0$ and $D_z$ in $z=0,y\le 0$.

The divergence of the polynomial vector field is $\nabla \cdot F=4xy$. Its volume integral over the enclosed quarter-ball vanishes by oddness in $x$. Thus the divergence theorem gives ${\Phi }_S+{\Phi }_{D_y}+{\Phi }_{D_z}=0$.

On $D_y$, the outward normal is $+\hat{y}$ and $F_y=-3z^4$. On $D_z$, it is $+\hat{z}$ and $F_z=x^3+y^2$; the odd term integrates to zero. The needed half-disc moments for radius $R$ areandAt $R=4$, the cap fluxes are $-768\pi$ and $32\pi$, respectively. Therefore the flux through a spherical quadrant by planar caps isThe source does not explicitly specify a normal for this part; the boxed answer adopts the usual outward spherical orientation. Choosing the inward normal reverses its sign.

Assume $a>0$ and choose the outward orientation. On the sphere, $\mathbf{n}=(x,y,z)/a$, so $\mathbf{F}\cdot \mathbf{n}=(x^2+y^2+2z^2)/a=(a^2+z^2)/a$. In spherical coordinates, $z=a\cos \theta$ and $dS=a^2\sin \theta d\theta d\varphi$. The direct surface integral isIndependently, $\nabla \cdot \mathbf{F}=1+1+2=4$. Its volume integral over the ball isThe two calculations agree exactly with the divergence theorem. The expression requested is a scalar flux of a vector field, namely the dot product with the directed surface area element.

Let $\mathbf{F}=(F_x,F_y,F_z)$ be continuously differentiable near $P$. Take a small axis-aligned rectangular volume around $P$, with side lengths $\Delta x,\Delta y,\Delta z$ shrinking with bounded aspect ratios. The outward flux through its two $x$ faces is, to leading order,where $\Delta V=\Delta x\Delta y\Delta z$. The analogous face pairs give $\partial F_y/\partial y$ and $\partial F_z/\partial z$. Dividing total flux by $\Delta V$ and taking the limit givesThe divergence is thus the local outward flux per unit volume; the symbols on the right are Cartesian partial derivatives.

For a bounded volume $V$ with piecewise smooth closed boundary $S$, subdivide into small cells. Each internal face occurs twice with opposite normal vectors, so its flux cancels. The remaining sum is the exterior surface integral; the cell sums of $(\nabla \cdot \mathbf{F})\Delta V$ converge to the volume integral. Assuming continuous first partial derivatives on a neighborhood of the closure of $V$, this proves the divergence theorem:The unit normal vector $\mathbf{n}$ points out of $V$. Multiple boundary components are allowed; at a cavity wall it points into the cavity.

For a continuously differentiable vector field $\mathbf{F}$ on a neighborhood of an oriented, piecewise smooth surface $S$, Stokes theorem statesHere $\partial S$ is the boundary curve, $d\mathbf{r}$ is its directed tangent element, $\hat{\mathbf{n}}$ is the chosen unit normal vector, $dS$ is the surface area element, and $\nabla \times \mathbf{F}$ is the curl. The right-hand rule relates the boundary direction to $\hat{\mathbf{n}}$. Thus circulation around the edge equals the flux of curl through the spanning surface. Reversing both orientations changes both signs. One can replace a difficult line integral by an easier surface integral, or replace one spanning surface by another within the region of regularity.

For a bounded volume $V$ with piecewise smooth closed boundary and a continuously differentiable vector field on a neighborhood of its closure, the divergence theorem statesHere ${\hat{\mathbf{n}}}_{\mathrm{out}}$ points out of $V$, including into any holes, $dV$ is the volume element, and $\nabla \cdot \mathbf{F}$ is the divergence. It equates total outward flux to integrated source strength. It converts a closed surface integral into a volume integral, and explains why internal fluxes cancel in a subdivided region. Singularities cannot be included without checking the hypotheses.

For a conservative force, the work around every closed path is zero. Stokes theorem gives zero flux of curl through every sufficiently small patch. Dividing by its area and shrinking the patch shows that every component of the continuous curl vanishes:Here $U$ is potential energy; conversely, zero curl implies a conservative vector field on a simply connected domain. The divergence theorem then gives the compatible relation ${\iint }_{\partial V}\mathbf{F}\cdot \hat{\mathbf{n}}dS=-{\iiint }_V{\nabla }^{2}UdV$. There is no general zero-divergence condition on a conservative force: $\mathbf{F}=-\mathbf{r}=-\nabla (r^2/2)$ is a concrete counterexample, with $\nabla \cdot \mathbf{F}=-3$.

Take $0<r_1<r_2$, so the shell excludes the possible singularity at the origin. The divergence and curl of a radial vector field formula in spherical coordinates givesWith volume element $dV=r^2\sin \theta drd\theta d\varphi$, direct integration givesFor the independent boundary calculation, the outer sphere has outward unit normal vector $+\hat{\mathbf{r}}$ and the inner sphere has outward unit normal vector $-\hat{\mathbf{r}}$. Their fluxes areTheir sum equals the direct volume integral, verifying the divergence theorem. The restriction $\beta \ne 2$ permits the displayed power-antiderivative step. At $\beta =2$, treated separately, both the shell volume integral and net boundary flux are zero; the excluded origin must still not be filled in.

Subtract the ambient velocity and stress, writing ${\mathbf{u}}^{\prime }=\mathbf{u}-{\mathbf{u}}_{\infty }$. This perturbation decays at infinity and on the sphere has velocity $\mathbf{U}+\mathbf{\Omega }\times \mathbf{x}-{\mathbf{u}}_{\infty }$. The ambient flow is regular through the ball and has no body force. Its net torque on the imaginary spherical surface is zero: the divergence theorem converts the torque to the volume integral of the stress divergence plus its antisymmetric part, both zero. Thus the perturbation has zero torque too.

Reciprocity with the auxiliary rotlet givesUsing ${\mathbf{\sigma }}_G{\mathbf{n}}_D=3(\mathbf{G}\times \mathbf{x})/(8\pi a^4)$ yieldsOn this sphere $\mathbf{x}=a\mathbf{n}$, so the vector divergence theorem gives $\int \mathbf{x}\times {\mathbf{u}}_{\infty }dS=a{\int }_{r<a}\nabla \times {\mathbf{u}}_{\infty }dV$. Taking the curl of the homogeneous Stokes equation makes ${\mathbf{\omega }}_{\infty }$ harmonic. Its ball average is its central value: the derivative of each spherical mean is proportional to the volume integral of its Laplacian, hence zero, and its small-radius limit is its value at the center. ThereforeThis proves Faxén rotation law directly from the reciprocal theorem, including why no finite-radius Laplacian correction is needed for a body-force-free ambient flow. Regularity of the ambient field throughout the sphere's prospective interior is part of this argument.

Away from the origin the gradients are $\nabla u=\hat{\mathbf{r}}/a$ and $\nabla v=-\hat{\mathbf{r}}/r^2$. The improper volume integral is thereforeThe function $v=1/r$ is a harmonic function only off the origin. It is singular inside the integration volume, so it does not meet part (a)'s regularity hypothesis. On the punctured ball, Green's first identity has outer flux $-4\pi$ and inner flux $4\pi \epsilon /a$: the inner outward unit normal is $-\hat{\mathbf{r}}$, so $u{\partial }_{n}v=(\epsilon /a)/{\epsilon }^2$. Their sum is $-4\pi +4\pi \epsilon /a$, exactly the integral before taking the limit. Equivalently the distributional Laplacian of $1/r$ is $-4\pi {\delta }_0$, rather than zero throughout the ball. This is boundary flux from a singular harmonic potential; it explains why applying part (a) across the origin would be invalid.

Here $v=z$, so $\nabla v={\mathbf{e}}_z$, while $\nabla u=\hat{\mathbf{r}}$ away from the origin. Their inner product is $\cos \theta$. The spherical coordinates volume element givesThis agrees with the boundary-constant gradient pairing with a boundary-constant function and a harmonic function: $u$ has constant boundary value $a$ and $v=z$ is a regular harmonic function. For comparison with the boundary value one in part (a), use $u/a$ and multiply back by $a$. Strictly, $u=r$ is not differentiable at the origin. Excising a ball of radius $\epsilon$ resolves this: the inner boundary term $u{\partial }_{n}v=-\epsilon \cos \theta$ has zero angular integral, so it contributes nothing in the limit. Thus the improper volume integral has the same zero value.