What Is the Magnetic Field Outside a Solenoid?

The magnetic field outside a solenoid depends on whether the solenoid is finite or idealized as infinitely long, and on where the field is measured. A finite solenoid produces a fringe field outside its windings. The familiar result that the outside field is zero applies to an infinitely long, ideal solenoid—not generally to a real one.

For a finite solenoid modeled as a continuous, uniformly wound current sheet, the simplest exact expression is for a point on its axis. At other points, the field can be calculated from the Biot–Savart law, but the result generally requires an integral or numerical evaluation.

Exact field on the axis of a finite solenoid

Model the solenoid as a thin cylindrical, continuous current sheet of radius $R$, length $L$, and turn density $n$ (turns per unit length), carrying current $I$. Put its center at $z=0$, so its ends are at $z=-L/2$ and $z=L/2$. The axial field at position $z$ is

$$B_z(z)=\frac{\mu_0 nI}{2}\left[\frac{z+L/2}{\sqrt{R^2+(z+L/2)^2}}-\frac{z-L/2}{\sqrt{R^2+(z-L/2)^2}}\right].$$

Here, $\mu_0$ is the permeability of free space. The sign of $B_z$ follows the chosen direction of current; the expression above assumes the current produces a field in the positive $z$ direction inside the solenoid. This result is exact for the continuous current-sheet model; for a real coil with discrete turns, it is an approximation unless the individual turns are summed.

This formula applies on the axis, including beyond either end. For a point past the right end, $z>L/2$, it gives the axial fringe field outside the solenoid. It also gives the familiar long-solenoid result near the center when $L$ is much greater than $R$ and the point is far from the ends: $B_z\approx\mu_0 nI$.

Worked example: a point beyond one end

Suppose a solenoid has radius $R=0.020\,\text{m}$, length $L=0.10\,\text{m}$, turn density $n=1000\,\text{m}^{-1}$, and current $I=2.0\,\text{A}$. Find the field on the axis at $z=0.070\,\text{m}$, which is $0.020\,\text{m}$ beyond the right end.

Substitute the values into the axial formula:

$$\begin{aligned} B_z&=\frac{\mu_0(1000)(2.0)}{2}\left[\frac{0.070+0.050}{\sqrt{0.020^2+0.120^2}}-\frac{0.070-0.050}{\sqrt{0.020^2+0.020^2}}\right]\\ &\approx 3.51\times10^{-4}\,\text{T}. \end{aligned}$$

So the field at this point is about $0.351\,\text{mT}$ along the solenoid’s axis. It is weaker than the approximate central field, $\mu_0 nI\approx2.51\,\text{mT}$, but it is not zero.

What if the point is off the axis?

For an off-axis point, the field depends on both its distance from the axis and its position along the solenoid. One exact way to specify it is the Biot–Savart integral for a continuous, tightly wound current sheet, at observation points not on the sheet. For an observation point $\mathbf r=(x,y,z)$, write each source point on the winding as $\mathbf r'=(R\cos\phi,R\sin\phi,z')$. Then

$$\mathbf B(\mathbf r)=\frac{\mu_0 nI}{4\pi}\int_{-L/2}^{L/2}\!\int_0^{2\pi} \frac{R(-\sin\phi,\cos\phi,0)\times(\mathbf r-\mathbf r')}{|\mathbf r-\mathbf r'|^3}\,d\phi\,dz'.$$

This expression adds the magnetic contributions from all the circular turns. It is exact for the continuous current-sheet model, but evaluating it at a general off-axis point usually calls for numerical integration. Closed forms involving elliptic integrals are also available for the field of a single circular loop; integrating those contributions along the solenoid’s length still leaves a more complicated expression than the on-axis result.

If the solenoid has a small number of distinct turns rather than a tightly wound, nearly continuous winding, model the turns individually. On the axis, for circular turns at positions $z_k$, the field is

$$B_z(z)=\sum_k\frac{\mu_0 I R^2}{2\left(R^2+(z-z_k)^2\right)^{3/2}}.$$

Why the infinite-solenoid answer differs

An infinitely long ideal solenoid has a uniform interior field and zero exterior field in the standard idealization. A finite solenoid has ends, so its field lines spread out near the ends and return outside. Making a solenoid longer makes the field near its middle more like the infinite-solenoid approximation; it does not make the field identically zero at every exterior point.

Common mistakes

• Using $B=0$ outside for a finite solenoid. That is the ideal infinite-solenoid result, not a general result for a finite coil.
• Applying the axial formula off axis. The formula above assumes the observation point lies on the axis. A point beside the cylindrical winding needs the off-axis field calculation.
• Omitting the observation location. “Outside” could mean beyond an end or beside the winding, and those locations do not generally have the same field.
• Confusing turns per length with total turns. For a uniform winding, $n=N/L$, where $N$ is the total number of turns.