Solenoid ($B={\mu }_{0}nI$)

A solenoid is a coil of electrical conductor, usually wire wound into a helix around a circular cylinder. The same electric current $I$ flows through its successive turns. Each turn contributes a magnetic field; the superposition principle makes their axial contributions reinforce inside the coil. Curl the fingers of your right hand in the direction of the electric current around the turns: the thumb gives the interior magnetic field direction by the right-hand rule. Throughout this branch, $N$ is the total number of turns, $\ell$ is the coil length, and $n=N/\ell$ is the solenoid turn density. ^[ref].

The attraction of a long solenoid is a controllable, nearly uniform magnetic field over an interior working region: change the electric current to tune its strength, or reverse the electric current to reverse its direction. This makes it a useful setting for studying the Lorentz force, measuring magnetic flux, and constructing an inductor. Its geometry turns the many contributions from individual turns into the simple long-solenoid approximation. A real finite coil has end fields and leakage; its exterior magnetic field is not zero. The on-axis magnetic field of a finite solenoid below quantifies that difference.

Historically, the coil helped connect magnetic fields to electric currents. In his memoir on the mathematical theory of electrodynamic phenomena, Ampère modeled the interactions of magnets using the ends of solenoids in place of microscopic magnetic constituents. This was a route toward explaining magnetism by electric currents, rather than a modern microscopic account of magnetization. ^[ref]. The historical setting is documented by the ^[ref].

For a near-axis beam of equal incoming speed $v$ through a uniform solenoid of length $l$, define $\omega =|qB|/(2\gamma m)$. Neglecting the small axial-speed changes, the internal radius is $\rho ={\rho }_0\cos (\omega z/v)$. The exit fringe removes the azimuthal velocity for zero incident canonical angular momentum. The field-free radial trajectory then reaches the axis a distancebeyond the exit, when $0<\omega l/v<\pi /2$. In the weak-lens limit, $f\simeq 4(\gamma mv{)}^2/(q^2{B}^{2}l)$. Without the paraxial approximation, the radius-dependent axial speed produces aberration; the ideal common focus is then only approximate.

Let a particle enter a uniform solenoid at transverse position $u_0=x_0+i{y}_0$ with zero radial velocity after an ideal fringe. Define signed $\Omega =qB/(2\gamma m)$ and initial field-free speed $v$. Its Lorentz force trajectory inside isThe transverse motion is a circle about the guiding centre $u_0/2$ at cyclotron angular frequency $2|\Omega |$. Cylindrical radius is $|u_0\cos \Omega t|$; its apparent azimuthal jumps when the orbit crosses the axis are coordinate singularities. Forward entry requires $v>|\Omega u_0|$ in this ideal model.

Near the axis of a solenoid, the magnetic vector potential $A_{\theta }=\rho B(z)/2$ givesThe radial fringe component enforces the solenoidal magnetic-field constraint. A sharp axial step in $B$ must retain this fringe as a distributional field; a purely axial field that jumps to zero violates Gauss's law for magnetism. The hard-edge model is a convenient limit for near-axis trajectories, not a literal monopole sheet.

The solenoid turn density $n$ counts turns per unit axial length in a uniformly wound solenoid. The total turn count $N$ is dimensionless, whereas $n$ has units ${\mathrm{m}}^{-1}$. A $1000$-turn coil of length $0.50\mathrm{m}$ has $n=2000{\mathrm{m}}^{-1}$; an axial interval of length $s=0.010\mathrm{m}$ then contains $ns=20$ turns. The effective circumferential surface current density in the closely spaced circular-turn model is $K=nI$, with units $A/m$. ^[ref].

For a single-layer helix, the axial advance per turn is approximately $1/n$. The circular-turn model neglects this advance and the associated axial component of electric current, as well as the detailed magnetic field close to individual wires. The number of turns around the whole coil is always $N=n\ell$ here; it must not be substituted for $n$ in $B={\mu }_{0}nI$.

In magnetostatics, an ideal infinitely long solenoid is a cylindrical sheet of circumferential surface current density $K=nI$. It has negligible winding advance, no magnetic core, and no imposed external magnetic field. Choose positive electric current to produce magnetic field along $+z$. In this ideal model the interior magnetic field is uniform and axial, $B={\mu }_{0}nI$, and the exterior magnetic field is zero; ${\mu }_0$ is the vacuum permeability. For a finite, closely wound air-core coil, the same result is an approximation well inside the coil, several radii from either end. Air is treated as having the magnetic permeability of vacuum. ^[ref].

For a short proof, the translational and rotational symmetries of the ideal cylindrical current sheet, together with the Biot-Savart law, give an axial magnetic field. Ampère's law on rectangles wholly outside makes that axial exterior value independent of radius; the condition of no imposed magnetic field makes it zero at infinity and hence everywhere outside. Rectangles wholly inside similarly show that the interior value is independent of radius. Now choose an Amperian loop with axial length $s$, one long side inside and the other outside. The transverse sides are perpendicular to the magnetic field, so their contributions to the line integral vanish. The loop crosses $ns$ turns of the winding, givingHere $s$ is the integration-path length, not the whole coil length $\ell$. An Amperian loop surrounding both sides of the winding encloses equal and opposite electric currents; zero net circulation alone would not prove zero magnetic field at every exterior point.

For the $1000$-turn, $0.50\mathrm{m}$ coil with electric current $I=0.50\mathrm{A}$, the long-solenoid approximation givesThis illustrative calculation uses vacuum permeability ${\mu }_0\approx 1.256637\times 10^{-6}H/m$, a rounded value, not an exact constant. Since the 2019 redefinition of SI, vacuum permeability is determined experimentally.^[ref] Doubling $I$ doubles the ideal magnetic field; doubling $N$ at fixed $\ell$ does the same. Increasing $\ell$ while keeping $N$ fixed decreases $n$ and hence the ideal magnetic field.

Model a finite solenoid as a continuum of coaxial turns shaped as circles of radius $a>0$, uniformly distributed over length $\ell$ with solenoid turn density $n=N/\ell$. Put its centre at $z=0$, so its ends are $z=\pm \ell /2$. With the same electric current orientation as in the long-solenoid approximation, its axial magnetic field isThis formula applies only on the axis, where symmetry cancels the transverse magnetic field. It does not describe points off the axis. It is exact for the continuum circular-turn model, which neglects winding thickness, discrete wire spacing, and helical advance. ^[ref].

The derivation is an especially direct use of the superposition principle. The on-axis magnetic field of a circular current loop at source position $z^{\prime }$ is ${\mu }_{0}I{a}^2/[2(a^2+(z-z^{\prime }{)}^2{)}^{3/2}]$. There are $nd{z}^{\prime }$ turns in an axial slice, soSubstitute $u=z-z^{\prime }$ and use the antiderivative $u/[a^2\sqrt{a^2+u^2}]$ to obtain the displayed expression. If the centre is instead at $z=z_0$, replace every $z$ in that expression by $z-z_0$.

At the centre and at either end, respectively,For aspect ratios $\ell /a=2,6,20$, the centre ratios are approximately $0.7071,0.9487,0.9950$. The end ratio tends to $1/2$ as $\ell /a$ becomes large, even though the centre ratio tends to $1$. Thus a long coil can have an excellent central magnetic field while the ends still differ strongly from that value. Convergence to the long-solenoid approximation is not uniform across the end region.

For a concrete geometry, give the earlier coil radius $a=0.025\mathrm{m}$, so $\ell /a=20$. Its central magnetic field is approximately $1.2504\mathrm{mT}$, its end value is $0.6275\mathrm{mT}$, and at $z=\ell$ its axial magnetic field is still $0.00277\mathrm{mT}$. The last point is outside the coil: finite length does not permit the zero-exterior conclusion of the infinite model. Off the axis, the magnetic field lines bend around to form the return field. A central axial value alone also does not determine the total magnetic flux through every turn of a short coil.

For a long vacuum-core or air-core solenoid with fixed geometry and linear magnetic response, neglecting end effects gives self-inductanceHere $N$ is the total number of turns, $n=N/\ell$ is the solenoid turn density, and $A$ is the interior cross-sectional area. The reason for the square of $N$ is physical: more turns produce more magnetic field, and more turns link the resulting magnetic flux. Approximately equal magnetic flux $\Phi \approx BA$ through each turn gives magnetic flux linkage $\Lambda \approx N\Phi \approx {\mu }_0{N}^{2}AI/\ell$; division by the electric current yields $L$. ^[ref].

For the same $N=1000$, $\ell =0.50\mathrm{m}$, and $a=0.025\mathrm{m}$, the area is $A=\pi (0.025{)}^2{\mathrm{m}}^2\approx 1.9635\times 10^{-3}{\mathrm{m}}^2$. The long-coil estimate isAt $I=0.50\mathrm{A}$, each turn links approximately $2.4674\times 10^{-6}\mathrm{Wb}$ and the total magnetic flux linkage is approximately $2.4674\times 10^{-3}\mathrm{Wb}$. Their ratio to $I$ is the stated self-inductance. A real finite coil need not give precisely these values: the axial centre correction above is not by itself a correction formula for inductance, which depends on magnetic flux across all the turns.

At fixed $A$ and $\ell$, doubling $N$ quadruples $L$. At fixed solenoid turn density $n$ and area $A$, doubling $\ell$ doubles both $N$ and $L$. These are different changes to the geometry, so there is no contradiction between $L\propto 1/\ell$ at fixed $N$ and $L\propto \ell$ at fixed $n$. A magnetic core requires its own magnetic permeability model; nonlinear magnetization can make a constant-$L$ description inadequate.

The magnetic energy stored in a linear solenoid of constant self-inductance $L$ isin the long-solenoid approximation. The first formula is an inductor relation; the second identifies its storage with magnetic energy in the interior magnetic field, under the vacuum-core approximation. This agreement joins the electrical circuit description to the spatial magnetic field description. The expression $B^2/(2{\mu }_0)$ is the vacuum magnetic energy density. ^[ref].

A short derivation starts from the terminal electric potential difference $v_L=LdI/dt$, with positive electric current entering the terminal at higher potential. The absorbed power is $v_{L}I$, and integrating from zero electric current givesThe result describes recoverable storage in the inductor; Joule heating in the wire is a separate expenditure of energy. ^[ref].

For the long-coil example, $L\approx 4.9348\mathrm{mH}$ and $I=0.50\mathrm{A}$ give $U\approx 6.1685\times 10^{-4}\mathrm{J}=0.61685\mathrm{mJ}$. Substituting the ideal $B\approx 1.256637\mathrm{mT}$ and volume $A\ell \approx 9.8175\times 10^{-4}{\mathrm{m}}^3$ into $B^{2}A\ell /(2{\mu }_0)$ gives the same value. Doubling electric current quadruples stored magnetic energy. In a real coil, reducing electric current can return this energy to an electrical circuit, or release it as Joule heating in a closed RL circuit.