Solenoid (physics)

A solenoid. The direct current i enters at the left and travels around the cylinder through the conducting wire. Two magnetic field lines are shown in blue. The magnetic induction B is a vector field tangent to field lines. The direction of B and i are connected via the right screw rule.

In physics, a solenoid is a spiral of insulating wire wound around a cylinder. When a direct current passes through the wire a magnetic field is generated. Usually the cylinder is made of iron or another ferromagnetic material that enhances the magnetic field.

The solenoid behaves like a (permanent) bar magnet with two poles: the North pole (the left side of the solenoid in the figure on the right) and the South pole (the right side of the depicted solenoid).

If the diameter of the cylinder is very much smaller than its length, the magnetic field H inside the solenoid, far from its left and right side, is homogeneous and of strength

$|\mathbf {H} |=Ni$

where N, the number of turns of wire per unit length, is assumed to be constant over the length of the cylinder. Further i is the amperage—the strength of the direct current running through the wire.

When the cylinder consists of a material with relative magnetic permeability μ_r then (in SI units) the magnetic induction (also often referred to as magnetic field) is,

$|\mathbf {B} |=\mu _{0}\mu _{\textrm {r}}|\mathbf {H} |,$

where μ₀ is the magnetic constant. Diamagnetic (e.g., gaseous nitrogen) and paramagnetic (e.g., gaseous oxygen) materials, have μ_r–values very close to unity. Ferromagnetic materials, like iron, have usually non-constant values that may vary, depending on |H|, by an order of magnitude. Materials with 10000 ≤ μ_r ≤ 50000 are known.

We see that the magnetic field H might be thought of as the magnetic field produced by the flow of current in the wire and the magnetic induction B as the total field, including also the contribution of the material inside the solenoid. The magnetizing field H may be weak, but the actual average magnetic field B inside the solenoid filled with ferromagnetic material maybe several orders of magnitude stronger, because B is greatly enhanced by the alignment of the material's natural atomic magnets in the direction of the field.

Magnetic field in infinitely long solenoid

By definition an "infinitely long" solenoid is so long that end effects are negligible for almost all of the solenoid. We will consider the middle part of a cylindrical solenoid where end effects are negligible, so that in effect the solenoid is infinitely long. In addition, we assume that the number of turns N per unit length is constant over the length. We will show consecutively that

• The magnetic field is parallel to the axis of the solenoid.
• The magnetic field does not vary over the length of the solenoid.
• The magnetic field does not vary across the diameter of the solenoid.
• The magnetic field is zero outside the solenoid.
• The strength of the homogeneous field inside the solenoid is given by |H| = Ni.

Field parallel to axis

Because of axial symmetry, the magnetic field of the solenoid is parallel to the cylindrical axis Z. In the figure below the field lines are given in blue. The fact that the field vectors are parallel to Z because of symmetry, is intuitively clear, but can be shown more explicitly.

To do this we refer, as usual, to the plane perpendicular to the Z-axis as the X-Y-plane (the diametric plane of the cylinder). One can show that the projections of all vectors B onto the X-Y plane cancel each other. This cancellation of components in the X-Y-plane has the consequence that only the components of the field vectors along Z contribute to the field, or in other words that the magnetic field is parallel to Z. To show this cancellation, we note that by symmetry a cylinder has infinitely many vertical mirror planes, i.e., mirror planes that pass through the Z axis. Consider now an arbitrary field vector B that is not parallel Z. Choose the orthogonal X and Y axis, spanning the X-Y plane, such that B is in the Z-Y plane. By assumption B has a non-zero component along the Y axis. The Z-X plane is one of the vertical mirror planes. Reflection of B in the Z-X plane gives the vector B', which is also in the Z-Y plane. The Y components of B and B' are of opposite sign and of the same length, so they cancel each other. Hence, by reflection symmetry every B has an image B' and the X-Y components of the two vecctors cancel each other.

Note finally that this argument holds for the field inside and outside the solenoid.

Field independent of Z

Consider in the figure the closed long narrow tube with top and bottom plane designated by A and B, respectively. Gauss' law reads:

$\iint _{\mathrm {closed} \atop \mathrm {surface} }\mathbf {H} \cdot d\mathbf {S} =0,$

where dS is a vector normal to the surface of the tube. The integral is zero for the long sides of the tube, because H and dS are orthogonal, and if the top and bottom plane are small enough (both equal to ΔS), the field may be taken constant over these planes, so that

$\iint _{\mathrm {closed} \atop \mathrm {surface} }\mathbf {H} \cdot d\mathbf {S} =|\Delta \mathbf {S} |{\big (}|\mathbf {H} _{A}|-|\mathbf {H} _{B}|{\big )}=0$

from which follows that the field strength |H_A| at A is the same as the strength |H_B| at B, and hence the field strength is independent of Z.

Note finally that this argument holds for the field inside and outside the solenoid.

Piece of "infinitely long" solenoid. Blue lines are magnetic force lines. Black dots are cross sections of current carrying wires. Red rectangles are explained in text.

Field constant across the diameter

Consider in the figure the rectangular closed path designated by 1, 2, 4, and 3. Ampere's law reads in SI units

$\oint _{C}\mathbf {H} \cdot d\mathbf {s} =i,$

where i is the electric current flowing through the closed path. In the path at hand i = 0. The vector ds is along the path. From orthonormality and |H| being constant along Z, follows that

${\begin{aligned}\oint _{C}\mathbf {H} \cdot d\mathbf {s} &=\int _{1}^{2}\mathbf {H} \cdot d\mathbf {s} +\int _{2}^{4}\mathbf {H} \cdot d\mathbf {s} +\int _{4}^{3}\mathbf {H} \cdot d\mathbf {s} +\int _{3}^{1}\mathbf {H} \cdot d\mathbf {s} \\&=|\mathbf {H} _{12}|s_{12}-|\mathbf {H} _{43}|s_{43}=0.\end{aligned}}$

Since the path lengths s₁₂ and s₄₃ are equal, it follows that the field strength |H₁₂| is the same as the strength |H₄₃|, and hence the field strength does not vary over the diameter.

Note finally that this argument holds for a closed path completely inside and completely outside the solenoid, the only condition being that no current flows through the closed path.

Magnetic field outside the solenoid

We just proved that the magnetic field outside the solenoid is constant, independent of the distance to the solenoid. Applying the physically plausible boundary condition that the field must go to zero for large distances, it follows that the field is equal to zero everywhere outside the solenoid.

Strength of the field inside the solenoid

Consider in the figure the rectangular closed path designated by a, b, d, and c. Ampere's law reads in SI units

$\oint _{C}\mathbf {H} \cdot d\mathbf {s} =LNi,$

where L N is the number of turns contained in the closed path. The total current through the path is LN times the amperage i. Along ac and bd the integral is zero because of orthogonality. Along cd the integral is zero because cd is outside the solenoid, where the field is zero. Along ab the field is constant (is H), hence

$\oint _{C}\mathbf {H} \cdot d\mathbf {s} =LH=LNi\quad \Longrightarrow \quad H=Ni.$