--- title: "The Biot-Savart Law and the Field of a Current Element" description: "For steady currents in vacuum, the Biot-Savart law adds vector field contributions from current elements. Each contribution depends on current, element length, angle and distance. Integration gives th" canonical: https://lightmysky.com/learn/science/the-biot-savart-law-and-the-field-of-a-current-element-mt_RRsHUaXrFq source: https://lightmysky.com/learn/science/the-biot-savart-law-and-the-field-of-a-current-element-mt_RRsHUaXrFq.md retrieved: 2026-11-12 --- > **Agent view.** This is the Markdown twin of the page, for tools and assistants. > When to use this site, and the call that answers each job: https://lightmysky.com/agent-instructions.md > API description (OpenAPI 3.1): https://lightmysky.com/openapi.json · Authentication: https://lightmysky.com/auth.md > Pricing: https://lightmysky.com/pricing.md · Catalog: https://lightmysky.com/llms.txt · Full catalog: https://lightmysky.com/llms-full.txt > Every machine-readable file on this domain: https://lightmysky.com/.well-known/ai-catalog.json > Ask for Markdown with `Accept: text/markdown`, a `.md` address, or `?mode=agent`. # The Biot-Savart Law and the Field of a Current Element For steady currents in vacuum, the Biot-Savart law adds vector field contributions from current elements. Each contribution depends on current, element length, angle and distance. Integration gives the fields of straight wires and circular loops. Subject: Science · Area: Electricity & Magnetism · Ages 19 to 20 Page: https://lightmysky.com/learn/science/the-biot-savart-law-and-the-field-of-a-current-element-mt_RRsHUaXrFq ## Ready when they can - Applies the Biot-Savart law to find the field of a straight wire and on the axis of a loop - Uses the cross product to fix the direction of each contribution - Finds the force between two parallel currents and says when it is attractive ## Lesson: Adding up magnetism piece by piece For steady current in a thin wire in vacuum, dB = (mu0 I/(4 pi r^2))(dl cross rhat), a vector equation. mu0 is vacuum permeability. dl follows conventional current; rhat is a unit vector from the element to the field point, a distance r away. Add contributions by vector integration. For a nonzero cross product, point your right index finger along dl and your middle finger along rhat; your thumb gives dB. The contribution's magnitude is dB = mu0 I dl sin(theta)/(4 pi r^2). Here dl is element length and theta is the angle between dl and rhat. For a nonzero contribution with current, length and angle fixed, doubling r divides dB by four. Parallel or antiparallel vectors give zero contribution. Treating a finite short section as one element is an approximation: distance and angle must vary very little across it. For an infinite straight wire, r is perpendicular distance to the wire and x measures position along it from the nearest point. Element distance is sqrt(x^2+r^2), and sin(theta) = r/sqrt(x^2+r^2). Here dx is a tiny wire length. An integral is the limit of sums as those lengths shrink to zero. At a fixed field point, all contributions share a direction, so B = (mu0 I/(4 pi)) integral from -infinity to infinity of r dx/(x^2+r^2)^(3/2). For r/(x^2+r^2)^(3/2), an antiderivative is x/[r sqrt(x^2+r^2)]. Its derivative gives the integrand. At the infinite limits, subtract lower from upper: 1/r minus (-1/r) = 2/r. Thus B = mu0 I/(2 pi r). At fixed current, doubling perpendicular distance halves B. Field lines circle the wire in perpendicular planes. Point your right thumb along current; your curled fingers follow B. At an arc's centre, dl is perpendicular to rhat and every contribution has the same axial direction. For radius R, use dl = R dphi. Integrating gives B = (mu0 I/(4 pi R)) integral from 0 to theta of dphi = mu0 I theta/(4 pi R). Theta must be in radians. A full loop has theta = 2 pi, giving B = mu0 I/(2R). Other circuit parts may add field. For a loop of radius R, x is axial distance from its centre. Split each field into components along the axis and perpendicular to it (transverse). Opposite elements cancel transverse components; axial components add. Each axial contribution is mu0 I R dl/[4 pi(R^2+x^2)^(3/2)]. Since integral dl = 2 pi R, B = mu0 I R^2/[2(R^2+x^2)^(3/2)]. At x = 0, B = mu0 I/(2R). Curl your right fingers with loop current; your thumb gives the axial field direction. Two long, thin, parallel wires in vacuum are separated by r. Wire 1 makes B1 = mu0 I1/(2 pi r) at wire 2, perpendicular to its current. The wire-force relation gives F = I2 L B1, so F/L = mu0 I1 I2/(2 pi r), using current magnitudes. Same-direction currents attract; opposite-direction currents repel. L is the length whose force is calculated. For a force-per-length comparison, constants cancel. Multiply the factors for I1 and I2, then divide by the separation factor. For example, tripling I1 while leaving I2 fixed and halving separation multiplies F/L by 3 divided by 1/2, or 6. Decide attraction or repulsion separately from the current directions. **Recap.** Add current-element fields as vectors. One fixed element follows 1/r^2; an infinite wire follows 1/r. Loop symmetry selects axial components. Parallel-current directions set attraction or repulsion. *Lesson, questions and flashcards written for Marble. Physics reference: University Physics Volume 2, sections 12.1 to 12.4.* ## Practice 8 questions on this page, each with its working shown. ## Needs first - [Magnetic Flux Density and the Force on a Current-Carrying Wire](https://lightmysky.com/learn/science/magnetic-flux-density-and-the-force-on-a-current-carrying-wire-mt_zuqc4gjBVh) ## Optional connections - [Kirchhoff's Rules for Multi-Loop Circuits](https://lightmysky.com/learn/science/kirchhoffs-rules-for-multi-loop-circuits-mt_AS3KiGstGD) - [Charged Particles Moving in a Magnetic Field](https://lightmysky.com/learn/science/charged-particles-moving-in-a-magnetic-field-mt_evuhZ6cSfI) ## Opens up - [Ampere's Law: Wires, Solenoids and Toroids](https://lightmysky.com/learn/science/amperes-law-wires-solenoids-and-toroids-mt_uTzRcSiUQ-)