Magnetic Field of an Infinite Straight Current-Carrying Wire
Calculate the field around a long straight wire with B=μ₀I/(2πa). Check units and direction, then work through a 20 μT example.
Syllabus path: Electromagnetism › Magnetic Field › Magnetic Flux Density
CSCA Exam Prep
This formula is aligned to the CSCA undergraduate admissions exam syllabus. Continue with the linked tutorial and topic practice.
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CSCA Physics Formula Reference
Formula and explanation
The field around a long straight wire: magnitude and direction
Direct answer: For an infinitely long, thin, straight wire carrying a steady current in vacuum, the field magnitude is B=μ₀I/(2πa). Here I is the current magnitude and a>0 is the perpendicular distance from the wire's axis. The field is tangent to a circle around that axis. For a real wire, use this approximation outside the wire, well away from its ends, with a small compared with its length. Using μ₀ is usually a good approximation in air.
How to calculate and describe the field
- Check the model: a long, straight wire, a steady current and a point outside the conductor. This expression does not directly describe the inside of a wire, the end of a short wire or the center of a coil.
- Measure a perpendicular to the wire's axis. Convert centimeters to meters and use amperes for the current.
- Substitute into B=μ₀I/(2πa), using μ₀≈4π×10⁻⁷ T·m/A for this calculation. The result is in teslas; 1 μT=10⁻⁶ T.
- Point your right thumb along conventional current and curl your fingers around the wire. A current coming out of the page toward you produces a counterclockwise field as you look at the page.
Worked example: 4 A at a distance of 4 cm
Assume the wire meets the conditions above and find the field produced by the wire itself. Convert a=4 cm to 0.04 m. Then B≈(4π×10⁻⁷×4)/(2π×0.04) T=2.0×10⁻⁵ T=20 μT. Changing one input at a time gives the comparisons below.
| Current I | Perpendicular distance a | Approximate field magnitude B |
|---|---|---|
| 4 A | 4 cm | 20 μT |
| 4 A | 8 cm | 10 μT |
| 8 A | 4 cm | 40 μT |
At fixed current, doubling the distance halves the field magnitude. At fixed distance, doubling the current doubles it. Reversing the current without changing its magnitude reverses the field direction, not its magnitude.
Checks before you finish
- Entering 4 instead of 0.04 for the distance in meters makes this answer 100 times too small.
- Using an inverse-square distance rule. This long-wire model gives B∝1/a, not 1/a².
- Giving only a positive number when the question asks for the magnetic field. State its direction too. For several wires, add their field contributions as vectors.
- Applying the expression near an iron core or to any wire shape. Different materials and geometries require an appropriate field model.
Formula reference and example notes
Last reviewed: . The reference explains the field model. The numerical example and comparison table are AICSCA practice material, not official exam questions.
Related tutorial and examples
Magnetic Flux Density
Continue with the magnetic flux density tutorial to distinguish field magnitude, field direction and magnetic force, and to choose a model that fits the question.
Plan your next CSCA study step
Use this topic in a focused practice set, then connect it to the exam requirements and timed papers.