Magnetic field of a current element
Derive the Biot-Savart current-element form and apply it to compute magnetic fields.
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Guided steps
- Keep the observation point fixed and subtract the source position from it. This displacement points from source to observation. Use its unit direction with distance squared, or the full displacement with distance cubed. Those are equivalent expressions, but their numerator and denominator must not be mixed.
- The moving carriers in a short source volume give charge times drift velocity equal to conventional current times directed length. For negative carriers, charge sign and drift direction already combine into the conventional current. There is no second electron reversal. The element remains part of a steady circuit.
- As the source moves along the wire, its distance to the fixed observation point and its direction can change. Keep those changes inside the vector integral. Parallel or antiparallel elements contribute zero at that point, but other elements can still contribute.
- For the wire on the x-axis and a point above it, the source-to-point displacement has a negative source-x component and a fixed positive y component. The x part drops out when crossed with the positive-x tangent. The surviving z contribution has a source distance that still changes with x.
- Only an effectively infinite straight source justifies infinite integration limits. Integrating the changing source geometry produces a field that falls inversely with perpendicular distance, rather than the inverse-square factor for one small element. A finite source needs its own limits and any neglected return path remains an assumption.
- The original long-wire calculation uses an already integrated result. Substitute perpendicular distance only after deciding that end and return effects are negligible. It is the full-wire field, not the contribution of an arbitrary short element.
- Now choose a short selected portion, not a long wire. Convert its length to metres, find the full source-centre separation, and take directed length cross displacement before calculating. Check whether the source geometry changes little across the selected length.
- Find the source-centre distance and signed z field contribution for the short selected portion. Keep the direction, approximation and selected-system explanation separate.
- Explain the cross-product direction, compare source length with distance to justify the approximation, and state why this is not the circuit total. Compare the argument yourself; two checked values do not mark those explanations.
- Test the source-length assumption. Keep the observation point fixed, write its displacement from a variable source coordinate, then integrate over the stated endpoints. The antiderivative is supplied so you can focus on the geometry, sign and correct limits.
- Calculate the selected finite portion’s signed z field contribution. Retain changing source distance and finite endpoints. Show and compare the definite integral separately.
- Show the definite integral and explain its sign, why the infinite-wire result differs, and how the finite result behaves for very long and very short sources. Compare these arguments yourself; the checker marks only the requested field component.
- Predict the selected element’s contribution for a parallel source tangent and source-to-point vector. Do not infer the whole circuit’s field from that one contribution.
- Calculate the field magnitude for the stated long-wire model. Choose perpendicular distance and keep the field distinct from a field per length.
- Predict the magnetic field when every steady current doubles with fixed directions and geometry. Apply linearity to the complete vector sum, including locations where that sum is zero.
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A current both experiences magnetic force and produces a magnetic field. The calculation changes its question: the force law uses the field at the wire, while Biot–Savart adds the field at an observation point from each source element. Keep the observation point fixed while moving along the source wire.
Use the moving-charge field model and vector cross products. Here currents are steady in thin wires in vacuum. A selected current element is part of a complete circuit, not an isolated piece with steady current appearing at one end and disappearing at the other. We may calculate one portion’s contribution without claiming it is the circuit’s total field.
Define the source-to-observation vector first
Let a source element be at r' and the observation point at rP. Define
The source-to-observation arrow ends at P. The directed element dℓ follows conventional current. With positive current magnitude I, the Biot–Savart law is
Use one form consistently: a unit vector needs R² in the denominator; a displacement vector needs R³. The field has units tesla because μ₀ has units T m/A and the remaining length ratio has units m⁻¹. For the numerical calculations here, use μ₀ ≈ 4π × 10⁻⁷ T m/A.
Relate the element to moving charge
For carrier number density n, signed charge q and cross-sectional area S, the current density is J = nqvd, where vd is drift velocity. In a small length dℓ, the carriers occupy volume S dℓ. Their moving-charge contributions therefore contain
If carriers are negative, both q and their drift direction reverse relative to conventional current; do not reverse the field a second time. This connection is a magnetostatic model for a steady current distribution, not a derivation of arbitrary time-dependent fields from a single instantaneous charge position. The field law and its source geometry are described in OpenStax’s Biot–Savart section.
Keep direction and distance inside the sum
The complete wire contribution is a vector integral:
The source coordinate changes along the integral. Consequently R, the source-to-P direction and the wire’s tangent can all change. Add signed components rather than the magnitudes of differently directed contributions. Several wires contribute through the same vector sum.
If φ is the angle between dℓ and R hat, then dB = (μ₀ I/4π)dℓ sin φ/R². At fixed element length, current and distance, the contribution is largest when those vectors are perpendicular. It is zero when parallel or antiparallel. This says nothing about contributions from other circuit elements. Reversing only current reverses the field contribution; doubling every current at fixed geometry doubles their vector sum.
A source position that changes along a straight wire
For a wire along the x-axis carrying current in + x and an observation point P = (0,a,0) with a > 0, write
The -xx hat part of R drops out of the cross product, giving
Every element of this particular straight portion gives + z field at P. That common direction is a geometrical result, not permission to add field magnitudes for every wire shape. The perpendicular distance a is fixed, but the source-to-P distance R varies with x.
Applying the integrated long-wire result
For an ideal straight wire extending far in both directions, use the limits x = -∞ to + ∞ in that element expression:
The full-wire field falls as 1/a, even though each element expression has an inverse-square distance factor. The integration changes how the contributing source lengths and angles combine. For a finite wire, keep the finite endpoints instead of borrowing the infinite limits. An unmodelled return path can contribute too; the long-wire approximation assumes its contribution and end effects are negligible at P.
Worked example 1
Field near a long wire
Problem
A long straight wire in vacuum carries 6.0 A. Find the magnetic-field magnitude at a perpendicular distance of 0.030 m, neglecting end effects. Use μ₀ ≈ 4π × 10⁻⁷ H m⁻¹.
Show full solution
Use the full-wire field
Method
Substitute the current and perpendicular distance in B = μ₀I/(2π a).Reason
This result already sums the contributions of all wire elements. A single-element 1/r² expression would give only one contribution.Working
Answer
B = ((4π × 10⁻⁷)(6.0))/(2π(0.030)) = 4.0 × 10⁻⁵ TState the field
Working
Answer
B = 4.0 × 10⁻⁵ T. The field is tangent to a circle centred on the wire, with sense given by the right-hand grip rule.
Quick check
At twice the distance the field halves, to 2.0 × 10⁻⁵ T: it falls as 1/a, not 1/a².
Approximate a short selected portion
Guided practice 1
Field from a short source
Problem
A straight portion of a complete steady-current circuit is centred at the origin, has length Δℓ = 0.200 mm and carries I = 3.0 A along + x. In vacuum find its approximate contribution Δ Bz at P = (0.030,0.040,0) m, using the centre’s source-to-P vector. First find the source-to-P distance. Explain the cross-product direction and why the approximation is justified. Use μ₀/(4π) ≈ 10⁻⁷ T m/A. Calculate this portion’s contribution only; the rest of the circuit is not included.
Write source position, observation position and their difference
Hints
Hint 1
Separate two distances
The perpendicular distance to the x-axis is not the full distance from the source centre to P.Hint 2
Choose one equivalent vector form
Use displacement divided by distance cubed, or the unit direction divided by distance squared. Compare source length with source-to-P distance.Show solution step by step
Build the displacement and check its variation
Method
Use R = (0.030x hat + 0.040y hat) m at the centre.Reason
The observation point is fixed; the selected source portion is short compared with this separation.Working
Answer
R = square root of (0.030² + 0.040²) = 0.050 m. The source length is 2.00 × 10⁻⁴ m, so Δℓ/R = 0.0040. Across this portion, using the central geometry is a good leading approximation.Evaluate the selected contribution
Method
Take Δℓ × R before its magnitude.Reason
The parallel x component contributes zero; x hat × y hat = z hat.Working
Answer
Δ Bz ≈ ((10⁻⁷)(3.0)(2.00 × 10⁻⁴)(0.040))/((0.050)³) = +1.9 × 10⁻⁸ T. This is not the total field of the complete circuit. Compare the cross product, small-length argument and system boundary yourself; the checker verifies only distance and field component.
Quick check
Keep finite source limits in a new geometry
Check your understanding 1
A straight portion of a complete steady-current circuit runs along the x-axis from x = -L to x = +L, with L = 0.030 m, and carries I = 4.0 A in + x. Find this portion’s contribution Bz at P = (0,a,0), with a = 0.040 m, in vacuum. Write the source-to-P vector and evaluate a definite integral. Compare your result with treating the portion as an infinite wire. Use μ₀/(4π) ≈ 10⁻⁷ T m/A. You may use ∫ a(x² + a²)^(-3/2)dx = x/(a square root of (x² + a²)). The rest of the circuit is outside this selected contribution.
Show hint
Show answer
At source coordinate x, R = (-x,a,0) and dℓ = dxx hat, so dℓ × R = a dxz hat. Therefore Bz = (μ₀ I)/(4π)∫_(-L)^L(a dx)/((x² + a²)^(3/2)); = ((μ₀ I)/(4π))2L/(a square root of (L² + a²)); = (10⁻⁷)(4.0)0.060/((0.040)(0.050)); = +1.2 × 10⁻⁵ T. The infinite-wire value is + 2.0 × 10⁻⁵ T: it includes extra source portions whose fields at P also point + z. The finite-to-infinite ratio is L/square root of (L² + a²) = 0.60. As L/a grows it approaches one; when L/a is small, the finite result tends to the centre-element approximation μ₀ I(2L)/(4π a²). Here 2L/a = 1.5, so that short-source approximation is not justified. Compare the definite integral, sign and limiting arguments yourself; checking one component does not mark them.
Check your understanding 2
What contribution does a current element make to the magnetic field at a point for which d vector l is parallel to r hat?
Show hint
Evaluate d vector l × r hat for parallel vectors.
Show answer
Its contribution is zero because sin 0° = 0, so d vector l × r hat = 0. Other elements of the circuit may still contribute at that point.
Check your understanding 3
A long straight wire in vacuum carries 2.0 A. Find the field magnitude at perpendicular distance a = 0.050 m using μ₀ ≈ 4π × 10⁻⁷ H m⁻¹.
Show hint
Use the integrated long-wire result, with a measured perpendicular to the wire.
Show answer
B = μ₀I/(2π a) = (2.0 × 10⁻⁷)(2.0)/0.050 = 8.0 × 10⁻⁶ T.
Check your understanding 4
If every steady current doubles without changing its direction or the conductor geometry, what happens to the magnetic field produced by those currents?
Show hint
Each element’s Biot-Savart contribution is proportional to its current.
Show answer
vector B doubles: every contribution doubles, so their vector sum doubles. The field direction remains the same wherever the original field is nonzero.
Build the field from directed source contributions
Fix the observation point, subtract each source position to form its displacement to P, take the cross product and add components over the actual source. Keep the selected portion distinct from the circuit, and decide whether a short-source or long-wire approximation is justified before using it.
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