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Biot-Savart vs Ampere's law, coil axis

Short answer

Biot-Savart law handles any current geometry by integrating dB = (μ₀/4π) I dl × r̂ / r²; Ampere’s law ∮B·dl = μ₀ I_enclosed requires symmetry. On the axis of a circular coil of N turns, radius R, current I, at distance x, B = μ₀ N I R² / [2 (R² + x²)^(3/2)].

Compare the methods of Biot and Savart Law and Ampere's Law to calculate magnetic fields due to current carrying conductor. Calculate magnetic field at an axial distance 'x' from the center of the circular coil carrying current.

Comparison

PointBiot-Savart lawAmpere's law
FormdB = (μ₀/4π) I dl × r̂ / r²∮B·dl = μ₀ I_enc
IdeaField of each small element, then add upLine integral of B round a closed loop
Symmetry neededNot neededNeeded (B constant along the loop)
UseAny shape of conductorInfinite wire, solenoid, toroid, sheet
MathsIntegration, often hardOften just algebra
AnalogyCoulomb's lawGauss's law

Ampere's law is quicker when symmetry exists. Biot-Savart works for every current distribution.

Field on the axis of a circular coil

Coil of radius R, N turns, current I. Point P is at distance x on the axis.

     I dl (element on the ring)
      o
      | \
    R |   \  r = √(R² + x²)
      |     \
      O-------P-----> x (axis)
          x
   (ring seen edge-on, centre O)
  1. Take a current element I dl on the ring. Its distance from P is r = √(R² + x²), and dl ⟂ r, so dB = (μ₀/4π) I dl / r².
  2. By symmetry the components of dB perpendicular to the axis cancel in pairs. Only the axial component of dB survives. It equals dB sin α, where sin α = R/r.
  3. dB_axial = dB sin α = (μ₀ I dl / 4π r²)(R/r).
  4. Integrate over the ring, ∮dl = 2πR: B = (μ₀ I R/4π r³)(2πR) = μ₀ I R²/(2 r³).
  5. For N turns:

B = μ₀ N I R² / [2 (R² + x²)^(3/2)]

At the centre (x = 0), B = μ₀NI/(2R). Far away (x ≫ R), B ≈ μ₀NIR²/(2x³).

✓ Answer checked by Claude Opus 5.5 (high effort)

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