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Parametrize the loop by $r^{\prime }=R{\hat{e}}_r(\varphi )$ and $d{\ell }^{\prime }=R{\hat{e}}_{\varphi }(\varphi )d\varphi$. At the observation point $r=z{\hat{e}}_z$, the Biot-Savart law for a line current givesThe components perpendicular to the axis cancel by rotational symmetry, while the axial component of the cross product is $R^{2}d\varphi$. ThereforeThe direction is fixed by the right-hand rule and reverses when the electric current reverses. At the centre,This is the on-axis magnetic field of a circular current loop.

For a field point in the plane of the loop, both the directed line element $d\mathbf{s}$ and the displacement $\mathbf{r}-\mathbf{s}$ lie in that plane. Their cross product is normal to it, so every term of the Biot-Savart law is normal, and their integral is normal as well. This argument also allows the field to vanish at particular points.

Choose the loop in the $xy$-plane with centre at the origin and let positive current be counterclockwise as seen from $+z$. On its lower side, write $\mathbf{s}=(t,-a,0)$, $-a\le t\le a$, soThat side contributesAll four contributions have the same direction and size by rotation. Thuswhere $\mathbf{n}$ follows the right-hand rule for the current.

Let the wire lie along $z$, and observe at perpendicular distance $r$. In the Biot-Savart law, the azimuthal contributions from line elements add with magnitudeIndependently, cylindrical symmetry makes the field constant and azimuthal on a circular Amperian loop, so Ampère's law gives $2\pi rB={\mu }_0{I}_0$. Both yieldwith direction fixed by the right-hand rule.

Choose the loop axis as $+x$, with positive electric current circulating by the right-hand rule. The Biot-Savart law isFor a loop element and an axial observation point, $r=\sqrt{a^2+x^2}$. The transverse components cancel between opposite elements. The axial component of $d\mathbf{\ell }\times \mathbf{r}$ is $ad\ell$, soThusReversing the electric current reverses the field. At the centre $B={\mu }_{0}I/(2a)$, while the far field decreases as $|x{|}^{-3}$, as expected for a magnetic dipole.

For a current element $Id\mathbf{\ell }$, the Biot-Savart law isTake a straight wire along $z$, extending from $-a$ to $a$, and observe its magnetic field at $(s,0,0)$. All contributions point in $+\hat{\mathbf{\varphi }}$ for positive $I$ along $+z$, by the right-hand rule. Their magnitude integrates toFor $a/s\to \infty$ this becomes ${\mu }_{0}I/(2\pi s)$, so $\oint \mathbf{B}\cdot d\mathbf{\ell }=2\pi s{B}_{\varphi }={\mu }_{0}I$, exactly as Ampère's law requires for an infinite straight wire.

Put the three wires at $(x,y)=(-d,0),(0,0),(d,0)$ with $d=1\mathrm{cm}$ and each current along $+z$, out of the page. Adding their Biot-Savart law magnetic fields givesFor $y\ne 0$, all three $x$-components have the same sign, so they cannot cancel. On $y=0$, a zero of the remaining component satisfiesExcluding the wire positions themselves, the two and only two finite zeros areThe magnetic field lines circulate counterclockwise about each current near that wire. Far away they circle the combined total current $3I$. The two nulls separate the nearby circulation regions in the original magnetic field line plot below.

For a steady filamentary electric current $I$, the Biot-Savart law isHere $\mathbf{r}$ is the observation position, ${\mathbf{r}}^{\prime }$ the source position, $d{\mathbf{\ell }}^{\prime }$ points in the direction of positive electric current, and ${\mu }_0$ is the vacuum permeability. A circular loop has centre magnetic field ${\mu }_{0}I/(2r)$: the cross product has magnitude $rd{\ell }^{\prime }$ and a constant axial direction, and $\int d{\ell }^{\prime }=2\pi r$.

Introduce the constant signed angular velocity $\Omega$ about the positive axis. A ring's electric charge $dq=2\pi \sigma rdr$ crosses any radial cut once per rotation, so its electric current is $dI=\Omega dq/(2\pi )=\sigma \Omega rdr$. Adding loop contributions yields the magnetic field of a rotating charged annulus:The direction follows the right-hand rule for positive electric charge; reversing either $Q$ or $\Omega$ reverses the magnetic field. The source supplies no rotation rate, so the answer necessarily retains $\Omega$.

The Biot-Savart law isHere $r^{\prime }$ is the current-element position, $d{\ell }^{\prime }$ follows the electric current, and $r$ is the observation point. For a circular current loop, transverse components cancel on the axis. Each element contributes axial factor $a/\sqrt{a^2+x^2}$ in addition to the inverse-square distance. Integrating around length $2\pi a$ gives the on-axis magnetic field of a circular current loopwith axial direction set by the right-hand rule.

For a finite straight wire parametrized by $z\in [-b,b]$, at perpendicular distance $a$, the Biot-Savart law givesThe direction is azimuthal about the current axis by the right-hand rule. In the hairpin, the semicircle contributes ${\mu }_{0}I/(4a)$ and each semi-infinite straight segment contributes ${\mu }_{0}I/(4\pi a)$. For the indicated current, all three contributions point out of the page. Thus the magnetic field at the center of a hairpin wire is

For a circular coil, use the Biot-Savart lawAt an axial position $x$, every wire element is at distance $r=\sqrt{a^2+x^2}$. The transverse magnetic field contributions cancel by rotational symmetry. The axial part of $d\mathbf{\ell }\times \mathbf{r}$ is $ad\ell$, and $\oint d\ell =2\pi a$. Multiplying by the $N$ turns givesThe electric current's sense fixes the axial direction by the right-hand rule.

For a circular current loop, the Biot-Savart law iswhere $\mathbf{s}$ points from a current element to the field point. Choose the loop in the $yz$ plane, with observation point $x\hat{\mathbf{x}}$. A parametrization consistent with positive axial magnetic field isThen $s=\sqrt{a^2+x^2}$ andThe transverse contributions cancel around the circle, leaving the on-axis magnetic field of a circular current loopThe right-hand rule fixes its direction; reversing the electric current reverses the magnetic field. At the centre this becomes ${\mu }_{0}I/(2a)$, and far away it decays as $|x{|}^{-3}$, as expected for a magnetic dipole.

Take the wires along $z$, with the first at $x=0$ carrying $+I\hat{\mathbf{z}}$, and observe at $x=-a$. The Biot-Savart law gives a first-wire field in $-\hat{\mathbf{y}}$; wire $n$ is $na$ away. ThereforePairing adjacent terms, with $N=2m$,Thus $\boxed{\mathbf{B}={\mu }_{0}IN(N+1)\hat{\mathbf{y}}/(4\pi a)}$. Its direction is opposite to the field of the first wire at the observation point. Reversing every electric current reverses this answer.

Pairing adjacent contributions from the Biot-Savart law gives magnitude ${\mu }_{0}IN(N+1)/(4\pi a)$, directed opposite to the first wire's field at the observation point.

Full derivation: solution.

For a steady thin-wire electric current, the Biot-Savart law iswhere the source element follows positive current. For a circular loop, source points are $(R\cos \varphi ,R\sin \varphi ,0)$ and the observer is $(0,0,z)$. Their cross product with $d{\mathbf{\ell }}^{\prime }$ has axial component $R^{2}d\varphi$; transverse components integrate to zero. Thus the on-axis magnetic field of a circular current loop, multiplied by the number of coincident turns, isTake positive current counterclockwise as seen from $+z$. magnetic field lines pass upward through the loop and return around its exterior, closing rather than terminating. They fan outward above its centre and converge below it. The original numerical Biot-Savart law integration below illustrates these directions; the wire singularities are masked.

The axisymmetric magnetic field near a regular axis follows from Gauss's law for magnetism. In cylindrical coordinates, regularity gives $B_{\rho }(0,z)=0$, while symmetry gives $B_z(\rho ,z)=B_0(z)+O({\rho }^2)$. Integrate ${\partial }_{\rho }(\rho B_{\rho })=-\rho B_0^{\prime }(z)$ from the axis:A formal $C(z)/\rho$ term is excluded by regularity. For $z>0$ the axial field decreases upward, so the radial component is outward, consistent with the sketch.

On the small receiving coil, $d\mathbf{\ell }=ad\varphi \hat{\mathbf{\varphi }}$. The Lorentz force $I_{2}d\mathbf{\ell }\times \mathbf{B}$ has axial component $-I_{2}a{B}_{\rho }(a,d)d\varphi$. Contributions from the axial field cancel radially around the loop. ThereforeWith both currents giving magnetic dipole moments along $+z$ and $d>0$, this is attraction. Reversing one current reverses the force. The small coil has $m_2=N_2{I}_2\pi a^2$, so the result is exactly the leading magnetic dipole force $F_z=m_2{B}_0^{\prime }(d)$. Equivalently $U=-{\mathbf{m}}_2\cdot {\mathbf{B}}_0$. The approximation requires $a\ll R$ and negligible variation over the receiving coil; corrections begin at relative order $a^2/(R^2+d^2)$.

Assume uniform charge per unit length and the same angular speed $\Omega$. A charge element at distance $r$ has $\mathbf{v}=\mathbf{\Omega }\times \mathbf{r}$, so the Biot-Savart law gives the centre field of a rigidly rotating charged loop as $dB={\mu }_0\Omega dq/(4\pi r)$. For one square side, with $b=a/2$,Using $dq=Qdx/(4a)$ on each side givesThus $B_{\square }/B_{\circ }=\ln (1+\sqrt{2})\simeq 0.8814$: the circular field is larger by $1/\ln (1+\sqrt{2})\simeq 1.135$. This is a rigidly rotating charge distribution, not a current constrained to the sides of a stationary square.

The Biot-Savart law gives the magnetic field of a steady thin-wire electric current in vacuum:The cross product fixes the circulation direction by the right-hand rule; fields from different electric current elements superpose. Integration gives $B={\mu }_{0}I/(2\pi r)$ outside a straight infinite wire and $B={\mu }_{0}I/(2a)$ at the centre of a circular loop of radius $a$. A distributed electric current replaces $Id{\mathbf{\ell }}^{\prime }$ by $\mathbf{J}d{V}^{\prime }$. This is a magnetostatic law: arbitrary time-dependent electric currents require retarded electromagnetic fields, not instantaneous substitution.

Take a long strip along $y$, occupying $-w/2<x<w/2$ in $z=0$, with uniform surface current $K=I/w$. Above the strip's centre, the contributions odd in $x$ cancel, while the Biot-Savart law gives the magnetic field of a uniformly conducting stripThe field reverses below the strip. Its field lines wrap around the longitudinal current: nearly parallel to the faces near the centre, then curving round the edges and approaching circles at large distances. The sketch is an original field integration; the conductor thickness is idealized as negligible.

Field law and centre value. The Biot-Savart law for a steady thin wire isHere $d{\mathbf{\ell }}^{\prime }$ follows the conventional electric current and the displacement points from source to observation point. For a circular loop of radius $a$, every element at the centre has distance $a$ and contributes in the same axial direction. Choose $+z$ by the right-hand rule. Then

On-axis integration. At axial distance $z$, every wire element has distance $\sqrt{a^2+z^2}$. Azimuthal symmetry cancels transverse components. Parameterizing ${\mathbf{r}}^{\prime }=(a\cos \phi ,a\sin \phi ,0)$ gives axial cross-product element $a^{2}d\phi$, so the on-axis magnetic field of a circular current loop isIt points the same way on the two halves of the axis; its strength is even in $z$.

Directions near the axial point. At a point on the positive axis the field is axial, but nearby lines spread outward because $B_0$ decreases as $z$ increases. On the negative-axis side they converge toward the axis while still progressing in the positive axial direction. There is no azimuthal field. The near-axis sketch shows this local spreading; the complete magnetic field lines return around the outside of the loop and form closed curves.

Radial component from zero divergence. Smooth axial symmetry gives $B_z(\rho ,z)=B_0(z)+O({\rho }^2)$ and $B_{\rho }=O(\rho )$. By Gauss's law for magnetism,Integrate radially from zero, excluding a singular $1/\rho$ term, to obtain the near-axis radial magnetic field, $B_{\rho }=-\rho B_0^{\prime }(z)/2+O({\rho }^3)$. Differentiating the axial field therefore givesFor $z>0$ it is outward; for $z<0$ it is inward. This supplies the direction reasoning used in the sketch.

Force on the second loop. Put the second loop at $z=d$ and radius $\rho =a$. For $d\gg a$, the radial field due to the first loop is $B_{\rho }\simeq 3{\mu }_{0}I{a}^3/(4d^4)$. The element $d\mathbf{\ell }=ad\phi \hat{\mathbf{\phi }}$ experiences Lorentz forceThe radial contributions cancel as vectors around the loop. The axial contributions add, giving the force between distant coaxial current loops,The sign follows $\hat{\mathbf{\phi }}\times \hat{\mathbf{\rho }}=-\hat{\mathbf{z}}$. Both the small radial coordinate relative to $d$ and the dipole approximation are justified by $a/d\ll 1$.

Force direction. The second loop is pulled back toward the first: the force is attractive for the same current sense. Reversing one current reverses the force. The much larger axial field chiefly produces radial element forces; retaining the smaller radial field is essential for the net axial attraction.

Use source coordinates $(x^{\prime },y^{\prime },0)$ and observation point $(0,0,z)$. The axial numerator in the Biot-Savart law isFor $|z|$ much larger than the loop dimensions, replace the denominator $[z^2+x^{\prime 2}+y^{\prime 2}{]}^{3/2}$ by $|z{|}^3$ to leading order. Transverse components cancel between opposite sides. The loop integral is twice its oriented area, soThereforeIt points along $\mathbf{m}$ on both halves of the dipole axis, not along the signed radial unit vector.

The field law. In the Biot-Savart law, $d\mathbf{B}$ is the contribution to the magnetic flux density at the observation point from a directed wire element $d\mathbf{\ell }$ carrying steady conventional electric current $I$. The vector $\mathbf{R}$ points from that element to the observation point, $R=|\mathbf{R}|$, and $\hat{\mathbf{R}}=\mathbf{R}/R$. The constant ${\mu }_0$ is the vacuum permeability; the cross product sets the direction by the right-hand rule. Integrating the elements gives the total field, in teslas.

The moment of the shown loop. The arrows in the original PDF run counterclockwise viewed from $+z$. Thus the magnetic dipole moment isIts units are $\mathrm{A}{\mathrm{m}}^2$. Reversing the current reverses both this moment and the field.

Magnetic field and force. A steady electric current produces a magnetic field. In vacuum the Biot-Savart law givesThe Lorentz force on the charge carriers transfers force to the material carrying them. The resulting force on a current element of another circuit is $d{\mathbf{F}}_2=I_{2}d{\mathbf{\ell }}_2\times {\mathbf{B}}_1$, or, for a continuous current distribution, the magnetic force per unit volume is $\mathbf{f}=\mathbf{J}\times \mathbf{B}$.

For two long parallel straight wires a distance $r$ apart, Ampère's law gives $B_1={\mu }_0{I}_1/(2\pi r)$. ThereforeThe force is attractive for parallel currents and repulsive for opposite currents, by the cross product rule. Its units are $\mathrm{N}{\mathrm{m}}^{-1}$. This approximation neglects wire thickness and end effects; a complete circuit includes its return current. Element-by-element forces need not form an isolated Newton's-third-law pair, but complete stationary circuits have balanced mutual forces when the entire system is included.

Energy description. For two linear coupled circuits the magnetic energy isHere $q$ is a mechanical coordinate and $M$ the signed mutual inductance. At prescribed currents, the generalized force is $I_1{I}_2\partial M/\partial q$ if the self-inductances do not vary. This positive derivative is obtained from magnetic coenergy, or by including the electrical sources' work; differentiating $W_B$ with a minus sign at fixed currents would omit that source work. At prescribed flux linkages the force instead equals the negative derivative of stored magnetic energy with those linkages fixed. The geometry thus determines both inductance and mechanical force.