Difficulty: Intermediate | Prerequisites: Ampere's law basics, right-hand rule, superposition of magnetic fields.
Infinite sheets of current are a standard Ampere's law geometry, much like infinite planes of charge are a standard Gauss's law geometry. The key insight is that each sheet produces a uniform magnetic field on either side, and you find the total field by superposing contributions from each sheet. This topic sits squarely in the "exploit symmetry" portion of the course and is a natural companion to the cylindrical-symmetry problems. If you can handle both cylindrical and planar Ampere's law problems, you have the two most common exam geometries covered.
Two infinite sheets of parallel wires create uniform magnetic fields that combine by superposition. The field from each sheet is B = ½ μ₀ n I, where n is the wire density and I is the current per wire. Use the right-hand rule to determine direction, then add the contributions from both sheets at any point.
Infinite current sheet
A flat surface carrying current uniformly distributed across its area. In practice, modelled as an infinite array of parallel wires with a given number of wires per unit length (n). Think of it as the magnetic equivalent of an infinite plane of charge: the field it produces is uniform and does not depend on distance from the sheet.
Wire density (n)
The number of wires per unit length across the sheet, measured perpendicular to the wire direction. If n = 18 wires/cm, convert to 1800 wires/m for SI calculations.
Surface current density (K)
The product n × I, giving the total current per unit length flowing across the sheet. In simple terms, it rolls up the wire density and per-wire current into a single number describing how much current flows per metre of sheet width.
Superposition (magnetic fields)
The total magnetic field at any point is the vector sum of the fields from each source. Each sheet contributes independently, and you add the contributions with proper attention to direction.
Two infinite sheets of current lie parallel to the y-z plane, equally spaced from the origin at x = −x₀ and x = +x₀, where x₀ = 4.2 cm.
Each sheet is an array of wires with density n = 18 wires/cm = 1800 wires/m.
Left sheet (x = −x₀): each wire carries I₁ = 3 A in the −z direction.
Right sheet (x = +x₀): each wire carries I₂ = 4.1 A in the +z direction.
By symmetry, the magnetic field from a sheet of wires flowing in the z-direction can only point in the y-direction (or −y).
The right-hand rule confirms: curl fingers in the direction of current (±z), and the field wraps around each wire in the x-y plane, but for an infinite sheet the x-components from neighbouring wires cancel. Only the y-component survives.
This applies to both sheets, so Bₓ = 0 everywhere.
Apply Ampere's law with a rectangular loop: one side parallel to the sheet on each side, two sides perpendicular.
The perpendicular sides contribute zero (B ⊥ dl).
The two parallel sides each have length L, and B is uniform and opposite in direction on each side of the sheet.
Ampere's law gives: B(2L) = μ₀ n L I, so B = ½ μ₀ n I on each side.
Direction: use the right-hand rule. For the left sheet (current in −z), the field points in −y for x > −x₀ and +y for x < −x₀. For the right sheet (current in +z), the field points in +y for x < +x₀ and −y for x > +x₀.
At point P, (x, y) = (−2.1 cm, 0), between the two sheets:
Left sheet contribution (B₁): current is in −z, point P is to the right of this sheet. Right-hand rule gives B₁ in the −y direction.
Right sheet contribution (B₂): current is in +z, point P is to the left of this sheet. Right-hand rule gives B₂ in the −y direction.
Both contributions point in the same direction (−y), so they add:
Bₚ = −½ μ₀ n (I₁ + I₂) = −½ (4π × 10⁻⁷)(1800)(3 + 4.1)
Bₚ = −0.00803 T
At point R, (x, y) = (−6.3 cm, 0), to the left of both sheets:
Left sheet: point is to the left, so B₁ is in the +y direction.
Right sheet: point is to the left, so B₂ is in the −y direction.
The fields partially cancel:
B_R = ½ μ₀ n (I₁ − I₂) = ½ (4π × 10⁻⁷)(1800)(3 − 4.1)
B_R = −0.00124 T
At point S, (x, y) = (6.3 cm, 0), to the right of both sheets:
Left sheet: point is to the right, so B₁ is in the −y direction.
Right sheet: point is to the right, so B₂ is in the +y direction.
Again the fields partially cancel, but in the opposite arrangement:
B_S = ½ μ₀ n (I₂ − I₁) = ½ (4π × 10⁻⁷)(1800)(4.1 − 3)
B_S = +0.00124 T
The integral ∮ B · dl around the trapezoid path a → b → c → d → a uses Ampere's law.
The loop encloses the left sheet's wires over the height H = 9.6 cm of the trapezoid.
I_enc = n × H × I₁ = 1800 × 0.096 × 3 = 518 A
By the right-hand rule (traversing the loop in the given direction), the enclosed current contributes negatively.
∮ B · dl = −μ₀ × 518 = −6.51 × 10⁻⁴ T·m
The B · dl dot product picks up only the component of B along the path direction.
Along the vertical segment from a to b (height h = 9.6 cm), the field is uniform (same distance from the sheets), so ∫ B · dl = B × h.
B at this location is the same as at point P (same x-coordinate), so:
∫_ab B · dl = (−0.00803)(0.096) = −7.71 × 10⁻⁴ T·m
Magnetic field from one infinite current sheet: B = ½ μ₀ n I
Superposition for two sheets (between the sheets, same-direction contributions): B_between = ½ μ₀ n (I₁ + I₂) (if both contributions point the same way)
Superposition for two sheets (outside both sheets): B_outside = ½ μ₀ n |I₁ − I₂| (partial cancellation)
Enclosed current for a finite section of an infinite sheet: I_enc = n × (length of sheet enclosed) × I_per_wire
Infinite current sheets are the idealised model behind Helmholtz coils and the field between the poles of large electromagnets. The uniform-field result is why MRI machines and particle accelerators use carefully designed planar or near-planar current distributions: when you want a uniform field over a large volume, a sheet geometry is your starting point.
Students often think the field from an infinite sheet depends on distance from the sheet. It does not. The field is uniform on each side, exactly like the electric field from an infinite plane of charge.
A frequent error is getting the direction wrong when superposing two sheets. Draw the right-hand rule for each sheet separately, at the specific point of interest, before adding.
Students sometimes forget to convert wire density from wires/cm to wires/m. This is a factor-of-100 error that shows up silently in the final answer.
When computing a closed-loop line integral, students occasionally forget that the sign of I_enc depends on the direction of traversal around the loop, not just the direction of the current.
⚠️ The formula B = ½ μ₀ n I for a single sheet is frequently required from memory. Know where the factor of ½ comes from (the rectangular Amperian loop has two contributing sides).
⚠️ Expect a question asking you to find the field in each of the three regions: left of both sheets, between them, and right of both sheets. The field between the sheets is typically the largest when currents flow in opposite directions (both contributions add).
⚠️ Sign errors on the line integral are the most common way to lose marks. Always state your loop direction and apply the right-hand rule to determine the sign of I_enc before computing.
True or false: The magnetic field from an infinite current sheet decreases with distance from the sheet. False. It is uniform on each side.
Fill in the blank: For a single infinite sheet with wire density n and current I per wire, B = ____. ½ μ₀ n I
True or false: Between two infinite sheets carrying currents in opposite z-directions, the magnetic fields from the two sheets always add. It depends on your position and the current directions. Between the sheets, the contributions can add or partially cancel depending on the specific arrangement. Work it out with the right-hand rule for the given problem.
Fill in the blank: To find the enclosed current for a line integral around a finite loop enclosing part of an infinite sheet, I_enc = n × ____ × I. The length (or height) of the sheet enclosed by the loop.
Q: Two infinite current sheets are parallel to the y-z plane at x = −5 cm and x = +5 cm. The left sheet has K = 200 A/m in the +z direction. The right sheet has K = 200 A/m in the −z direction. What is Bᵧ at the origin?
A: Between the sheets, both contributions point in the same direction (work out with right-hand rule). B = ½ μ₀ K + ½ μ₀ K = μ₀ K = (4π × 10⁻⁷)(200) = 2.51 × 10⁻⁴ T. The sign depends on the directions and the right-hand rule.
Q: In the same setup, what is Bᵧ at x = +10 cm (to the right of both sheets)?
A: The two contributions are in opposite directions and equal in magnitude, so they cancel: Bᵧ = 0. (This is analogous to the electric field outside a parallel-plate capacitor.)
Q: An infinite current sheet with n = 500 wires/m, each carrying 2 A in the +z direction, lies in the y-z plane. What is the magnitude of B at any point to the right of the sheet?
A: B = ½ μ₀ n I = ½ (4π × 10⁻⁷)(500)(2) = 6.28 × 10⁻⁴ T.
This connects to the parallel-plate capacitor analogy in electrostatics: just as two charged planes create a uniform E field between them and zero outside (for equal and opposite charges), two current sheets can create a uniform B field between them. It also connects to solenoids, which can be thought of as many current sheets stacked together, producing a strong, uniform interior field.
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