Ampere's Law: Wires, Solenoids and Toroids
For steady currents in vacuum, the closed-path integral of B dot dl equals mu0 times signed enclosed current. Sufficient symmetry lets this determine fields. An ideal infinite solenoid has a uniform interior field and zero exterior field.
What a learner can do afterwards
- States Ampere's law and chooses a loop that matches the symmetry of the current
- Derives the field of a long straight wire, a solenoid and a toroid
- Explains why Ampere's law constrains circulation but usually cannot determine a local field without sufficient symmetry
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Biot-Savart adds fields from current elements. For steady currents in vacuum, Ampere's law instead relates circulation to current: integral around a closed path of B dot dl = mu0 Ienc. The dot product selects the field component along each path element. Ienc is the signed current crossing a surface bounded by the path. This equation gives circulation, not automatically the field at each point.
Curl your right fingers in the chosen path direction; your thumb gives the positive surface normal. Crossings with it count positive and crossings against it count negative. For example, +4 A and -1 A give Ienc = 3 A. External currents can affect B on the path, although their contribution to its circulation is zero. Zero net enclosed current does not by itself mean zero local field.
For an infinite straight wire, choose a circle centred on the wire in a perpendicular plane. Cylindrical symmetry makes B tangent to the circle and constant in magnitude. Thus integral B dot dl = B(2 pi r) = mu0 I, giving B = mu0 I/(2 pi r). This agrees with Biot-Savart. A long real wire approximates this result away from its ends.
An ideal infinite solenoid has uniformly distributed turns, with vacuum inside. Symmetry and the condition that the field vanishes far outside give zero exterior field and a uniform axial interior field. A finite, tightly wound solenoid approximates this near its middle when its length greatly exceeds its radius. Its exterior field need not be zero.
These bars compare magnitudes in the ideal model; they do not draw field lines.
For that ideal solenoid, use a rectangle with an axial side of length l inside and a matching side outside. The outside field is zero; the end segments are perpendicular to B. Only the inside side contributes B l. With n = N/L turns per metre, the surface crosses n l turns, so B l = mu0 n l I and B = mu0 n I. At fixed model conditions, field magnitude scales with n times I.
For an ideal toroid, N circular turns are spread uniformly around a ring with vacuum in its winding region. At radius r from its central axis within that region, symmetry makes B tangent and constant on a centred circle. Its surface crosses N turns: B(2 pi r) = mu0 N I, so B = mu0 N I/(2 pi r). Symmetry also gives zero field in the hole and exterior. Real windings and leads can produce leakage.
An infinite circular wire of radius a carries total current I uniformly across its cross-section. A centred circle of radius r <= a encloses current in proportion to area: Ienc/I = pi r^2/(pi a^2) = (r/a)^2. Then B(2 pi r) = mu0 I(r/a)^2, so B = mu0 I r/(2 pi a^2) inside. Outside, all I is enclosed and B = mu0 I/(2 pi r).
The bars show current shares, not a drawing of the wire's cross-section.
Without enough symmetry, Ampere's law still determines circulation from signed current. Finding a local B directly requires a path where its tangential component and the other contributions are known. A single circular current loop has rotational symmetry, but no useful path isolates its axial field as B times a known length. Use the Biot-Savart axial calculation for that case. The circulation law still applies.
For steady currents in vacuum, circulation equals mu0 times signed enclosed current. Use symmetry to find local fields, and keep the ideal wire, solenoid and toroid assumptions explicit.
Original lesson revisions, questions and flashcards written for Marble. Physics facts and calculation methods were studied using University Physics Volume 2, sections 12.5 and 12.6, and College Physics, section 22.9; textbook prose is not reproduced.
Readings and flashcards written for Marble. Physics references: University Physics Volume 2, sections 12.5 and 12.6, and College Physics, section 22.9. Video links are credited to Khan Academy; their licenses are unknown.
More to read
A zero integral can contain nonzero fields
An Amperian path is an imagined closed route. At each point, B dot dl selects the component of magnetic field along the route. The closed-path integral adds these signed contributions.
For steady currents in vacuum, this circulation equals mu0 Ienc. Choose the positive surface normal with the right-hand rule: fingers follow the path direction and the thumb defines positive current crossings. Opposite crossings subtract.
Knowing the circulation does not usually determine every local field value. A path with zero enclosed current has zero circulation, yet B can be nonzero along it. Contributions can cancel, or the field can be perpendicular to parts of the path. External currents can affect those local values while contributing zero circulation.
For an infinite straight wire, cylindrical symmetry supplies the missing information. On a centred perpendicular circle, B is tangent and constant in magnitude. The integral becomes B(2 pi r), which can be solved for B.
Separate the coil from its ideal model
The infinite-solenoid model replaces discrete turns by a uniform winding distribution. With vacuum inside, its field is uniform and axial in the interior and zero outside. A rectangular path then has one contributing axial side: B l = mu0 n l I, giving B = mu0 n I. Here n = N/L is turns per metre.
A finite solenoid does not have exact infinite-solenoid symmetry. The formula approximates its interior near the middle when coil length greatly exceeds radius. Near the ends, field size and direction differ, and an exterior field can remain.
An ideal toroid distributes circular turns uniformly around a ring. Within its vacuum winding region, a circle centred on the ring's axis encloses N current crossings. Its field is tangent, so B(2 pi r) = mu0 N I. The resulting B = mu0 N I/(2 pi r) varies with distance r from that axis. The ideal hole and exterior have zero field. Real winding details and supply leads can produce leakage.
Inside a current-carrying wire
An infinitely long circular wire of radius a carries total current I. Suppose current density is uniform over its cross-section. A centred circle of radius r inside the wire encloses a current share equal to its area share.
That share is pi r^2/(pi a^2) = (r/a)^2. Substituting Ienc = I(r/a)^2 into Ampere's law gives B(2 pi r) = mu0 I(r/a)^2. Hence B = mu0 I r/(2 pi a^2) inside the wire.
The interior field rises linearly from zero on the axis. At the surface it reaches mu0 I/(2 pi a). Beyond the surface, the enclosed current stops growing because the whole wire current is already included. The field then decreases as mu0 I/(2 pi r).
The area-share method depends on uniform current density. If the distribution changes with radius, calculate enclosed current from that distribution instead.
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