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If coaxial loops of radii $R$ and $2R$, separated by $D$, carry currents $I$ and $2I$ in opposite senses, their on-axis fields cancel between the loops at distance $D/3$ from the smaller loop.

A circular loop of radius $R$ carrying electric current $I$ has, on its axis,where the sign of $\hat{z}$ is fixed by the right-hand rule. At its centre, $B(0)={\mu }_{0}I/(2R)$.

Put the loop in the $xy$ plane, centered at the origin, with positive electric current counterclockwise when viewed from $+z$. The right-hand rule then gives the central magnetic field along $+z$. Parameterize source positions by the angle $\varphi$ and observe at $(0,0,z)$:Here $R$ without bold type is the loop radius, whereas $\mathbf{R}$ is the source-to-observer displacement. Their cross product isThe denominator $|\mathbf{R}{|}^3=(R^2+z^2{)}^{3/2}$ is constant around the loop. The integrals of sine and cosine over a full turn vanish, while the axial contribution adds to $2\pi R^2$. Substitution into the Biot-Savart law proves the on-axis formula. Symmetry cancels transverse components; it does not cancel the axial components.

For an original one-turn example in the same vacuum approximation, choose $R=0.04\mathrm{m}$ and $I=2\mathrm{A}$. With ${\mu }_0\approx 4\pi \times 10^{-7}\mathrm{T}m/A$,At $z=0.03\mathrm{m}$, the source-to-observer distance is $0.05\mathrm{m}$, so $B(z)/B(0)=64/125=0.512$ and $B(z)\approx 16.08\mu \mathrm{T}$. At $z=-0.03\mathrm{m}$, the magnetic field has the same magnitude and still points along $+z$. The point has crossed the loop plane, but the loop's electric current has not reversed. Reversing the electric current reverses the magnetic field on both sides.

For $|z|\ll R$, a Taylor expansion gives $B(z)/B(0)=1-3z^2/(2R^2)+O(z^4/R^4)$: the center is a smooth maximum, with zero first derivative. For $|z|\gg R$,This is the axial magnetic dipole field of magnetic dipole moment $\mathbf{m}=I\pi R^2\hat{\mathbf{z}}$.^[ref] A compact closed loop has an inverse-cube axial far field, in contrast to the inverse-distance field of the infinite straight wire. The SI units of the exact loop formula are ${\mu }_{0}I{R}^2/(R^2+z^2{)}^{3/2}\sim {\mu }_{0}I/\text{length}$, hence tesla.

For $N$ coincident turns carrying the same electric current, the superposition principle multiplies the ideal result by $N$. A real coil has radial width and axial length: if those are not negligible, integrate or sum over its individual turns. This is why a single measured coil diameter and a turn count may not determine the full spatial magnetic field precisely. A collection of loops becomes a solenoid; integrating their axial contributions gives the on-axis magnetic field of a finite solenoid.

A pinned filament of length $L$ and bending modulus $A$ first buckles under compressive force $F_c={\pi }^{2}A/L^2$.

The worm-like chain models an inextensible filament with bending energy $A\int {\kappa }^{2}ds/2$ and thermal tangent fluctuations.

Resistive-force theory approximates the local viscous force on a moving slender elastic filament by separate drag coefficients along and across its tangent. Let $\mathbf{t}$ be the unit tangent and $\mathbf{v}$ the material velocity relative to the ambient fluid. The force per unit length on the elastic filament isThe positive parallel and perpendicular drag coefficients of a slender filament are defined per unit length; for a sufficiently slender isolated body, their leading logarithmic ratio is ${\xi }_{\perp }/{\xi }_{\parallel }\simeq 2$.

The directional difference can generate propulsion. For example, take a straight element with $\mathbf{t}=(\cos \theta ,\sin \theta )$ and impose transverse velocity $\mathbf{v}=V{\mathbf{e}}_y$. Direct substitution givesA transverse motion of a tilted element therefore has an axial force component. In a propagating flagellum deformation, the correlation of tilt and transverse velocity can produce nonzero mean thrust. For a freely swimming body, the unknown translation and rotation follow by imposing zero total external force and torque, including the resistance of the head if present.

This is a local Stokes flow approximation. It neglects the nonlocal hydrodynamic interactions retained by slender-body theory, so nearby boundary conditions, tightly spaced elastic filaments and detailed end effects can require a more complete model.

Cytoplasmic streaming is directed cytoplasmic flow within a cell, often driven by molecular motors acting along the boundary and used to enhance intracellular transport.

Microcirculation is blood flow through arterioles, capillaries, and venules, where vessel dimensions are comparable with cellular length scales.

For a nonnegative integer $n$ and commuting quantities $x,y$,In expanding the $n$ factors, obtain $x^k{y}^{n-k}$ by selecting the $k$ positions that contribute $x$. There are $\left(\genfrac{}{}{0ex}{}{n}{k}\right)$ such choices. For example,The coefficients are row five of Pascal's triangle, in reverse order under the displayed $x^k{y}^{n-k}$ convention; the row is symmetric. Setting $x=y=1$ gives $1+5+10+10+5+1=32$, the number of subsets of a five-element set. Each member has two inclusion choices, agreeing with the rule of product and the cardinality of the power set.

For nonnegative integers with $n_1+\cdots +n_r=n$, the multinomial coefficientcounts arrangements of a multiset containing $n_j$ copies of symbol $j$.

The central binomial coefficientcounts the $n$-element subsets of a $2n$-element set. It is the middle and largest coefficient in the binomial expansion of $(1+x{)}^{2n}$, by unimodality of binomial coefficients. For $n=2$, the row $1,4,6,4,1$ has central value six.

Equivalently, $\left(\genfrac{}{}{0ex}{}{2n}{n}\right)$ counts binary strings of length $2n$ with exactly $n$ ones. Dividing by the total $2^{2n}$ gives the probability of exactly $n$ successes in $2n$ independent fair Bernoulli trials. The normalized central binomial coefficients therefore connect a simple counting sequence to balanced random outcomes.

At a fixed positive integer level $r$, the combinatorial number system writes each positive integer $m$ uniquely asChoose $a_r$ as large as possible with $\left(\genfrac{}{}{0ex}{}{a_r}{r}\right)\le m$, subtract that term and repeat at level $r-1$, stopping when the remainder is zero. For example, at level three,Pascal's identity gives $\left(\genfrac{}{}{0ex}{}{a+1}{j}\right)-\left(\genfrac{}{}{0ex}{}{a}{j}\right)=\left(\genfrac{}{}{0ex}{}{a}{j-1}\right)$, so the remainder after choosing the level-$j$ term is smaller than $\left(\genfrac{}{}{0ex}{}{a}{j-1}\right)$. The next top index is consequently strictly smaller. Hockey-stick identity also bounds the sum of all lower admissible terms by $\left(\genfrac{}{}{0ex}{}{a}{j-1}\right)-1$. Thus those terms cannot replace a larger level-$j$ choice, proving uniqueness. The zero integer has the empty representation. This correspondence is useful for ranking and unranking fixed-size subsets.

The identitycomes from an integer change-of-variable matrix of determinant $-1$. It therefore preserves the set of integers represented and reduces the form to $u^2+v^2$.

An integral binary quadratic form $[a,b,c]$ with $a>0$ and negative discriminant of a binary quadratic form is reduced whenwith the additional convention $b\ge 0$ whenever $|b|=a$ or $a=c$. These inequalities bound the polynomial coefficients without changing the represented integers.

Every proper equivalence class of positive definite forms has exactly one reduced representative with this boundary convention. For primitive binary quadratic forms, this is the classical reduction theorem; the general case follows by dividing by the common content, which is invariant under proper equivalence.^[ref]

The sign rule removes actual duplicates. A shear takes $[a,a,c]$ to $[a,-a,c]$, and the determinant-one rotation $(x,y)\mapsto (-y,x)$ takes $[a,b,a]$ to $[a,-b,a]$. Keeping the nonnegative middle polynomial coefficient chooses one boundary copy. For example, $[1,2,2]$ is not reduced; the reduction algorithm for a positive definite binary quadratic form takes it back to $[1,0,1]$.

The discriminant of a binary quadratic form $[a,b,c]$ is $D=b^2-4ac$. For integer polynomial coefficients, $D$ is congruent to zero or one modulo four. Completing the square givesThus for real polynomial coefficients with $a>0$, positive definiteness holds exactly when $D<0$. For $D>0$ the form is an indefinite quadratic form, taking values of both signs; for $D=0$ it is degenerate. For example, $x^2+y^2$ has $D=-4$, whereas $x^2-y^2$ has $D=4$.

Over the real numbers, use the symmetric matrix $A=\left(\begin{array}{cc}a& b/2\\ b/2& c\end{array}\right)$. Its determinant is $-D/4$. An integer coordinate change $M$ gives $A^{\prime }=M^{T}AM$, soIn particular unimodular matrices preserve $D$, proving its invariance under proper equivalence. Changes with other determinants can change the discriminant. For instance, replacing $x$ by $2x$ takes $x^2+y^2$ to $4x^2+y^2$ and multiplies $D$ by four. The chosen half-coefficient matrix convention matters in this calculation.

An integral binary quadratic form $[a,b,c]$ is primitive when $\gcd (a,b,c)=1$. This condition concerns its polynomial coefficients, rather than the values of a particular representing pair. For example, $[1,0,3]$ is primitive, while $[2,0,4]$ has a common divisor two and represents only even integers.

Primitivity is invariant under proper equivalence. Expanding an integer change of variables expresses every new polynomial coefficient as an integer linear combination of the old ones, so every common divisor of the old coefficients divides the new ones. The integer inverse gives the converse. Their greatest common divisor, called the content, is therefore unchanged.

A primitive representation separately means $\gcd (x,y)=1$ in $n=f(x,y)$. Even a primitive form can have nonprimitive representations: $20=2^2+4^2$ for $[1,0,1]$. A representation of a prime number, however, must use a coprime pair, since a common divisor $d>1$ would force $d^2$ to divide that prime. This observation drives the discriminant criterion for prime representation by a binary quadratic form.

Integral binary quadratic forms $f$ and $g$ are properly equivalent ifThe matrix $M$ has integer entries and determinant one. Its inverse $\left(\begin{array}{cc}\delta & -\beta \\ -\gamma & \alpha \end{array}\right)$ also has integer entries, so the change is a bijection of the integer lattice. Consequently $f$ and $g$ represent the same integers, and the discriminant of a binary quadratic form is unchanged.

Allowing determinant minus one as well gives the broader equivalence under unimodular matrices. Proper equivalence retains orientation. For example, the shear $(x,y)\mapsto (x+y,y)$ is proper and takes $[1,0,1]$ to $[1,2,2]$. The reflection $(x,y)\mapsto (x,-y)$ instead has determinant minus one and takes $[a,b,c]$ to $[a,-b,c]$; those two forms need not be properly equivalent. This distinction matters when counting the class number of a negative discriminant.

If almost all neutrons enter helium-4 and $r=n_n/n_p$, its primordial mass fraction is $Y_p\simeq 2r/(1+r)$.

Light nuclei cannot accumulate until cooling makes deuterium sufficiently resistant to photodissociation, delaying nucleosynthesis after weak freeze-out.

Weak neutron--proton conversion freezes out when its interaction rate falls below the Hubble expansion rate.

With negligible lepton chemical potentials, weak equilibrium gives $n_n/n_p\simeq e^{-Q/(k_{B}T)}$, where $Q=(m_n-m_p)c^2$.