Biot–Savart lab

What it does

The Biot–Savart law says that every short piece of a current-carrying wire makes a little magnetic field around it: dB = (μ₀/4π) I dl sin θ / r². To find the field of a whole wire, you add up the pieces. This page lets you do that adding, one piece at a time, for three shapes:

• Straight wire: every piece pushes the field the same way (into the page at P), so the pieces simply add. A bar chart shows that the pieces nearest P give the most.
• Circular loop: each piece’s field at a point on the axis is tilted. Pieces on opposite sides cancel each other’s sideways parts, leaving only the part along the axis. You see each piece’s arrow and the running total.
• Solenoid: many loops side by side. The graph shows the field along the axis: flat at μ₀nI inside a long solenoid, falling to about half at each end.

The fourth mode is Ampère’s law. Three long wires come out of or go into the page. Drag a loop among them: the field along the loop changes, but the total ∮B·dl is always μ₀ times the current inside.

It follows the Class 12 Physics chapter on moving charges and magnetism (the Biot–Savart law, the field on the axis of a circular loop, Ampère’s circuital law, the solenoid). See the field lines with the magnetic field mapper, then use the field on a moving charge with lorentz motion or build a real coil with the electromagnet builder.

How to use it

1. Choose Straight wire, Circular loop, Solenoid or Ampère’s law.
2. Set the current and the sizes with the sliders. P is the point where the field is found.
3. For the wire and loop, slide Pieces of wire added, or press Add the pieces one by one to watch the total grow.
4. In Ampère’s law, set each wire’s current (positive is out of the page) and move the loop with the sliders or by dragging it on the picture.

Worked example

A straight wire 20 cm long carries 5 A. At P, 5 cm from the middle of the wire, sin α₁ = sin α₂ = 10/√(5² + 10²) = 0.894, so B = μ₀I/(4πd) × (0.894 + 0.894) = 10⁻⁷ × 5 / 0.05 × 1.789 = 17.9 µT, into the page. Adding the 24 pieces gives the same to within 0.02%. A very long wire would give μ₀I/(2πd) = 20.0 µT, so this short wire gives 89.4% of that.

A single loop of radius 10 cm with 5 A gives μ₀I/(2R) = 31.4 µT at its centre. A solenoid 20 cm long with 200 turns of radius 2 cm and 5 A has n = 1000 turns per metre, so μ₀nI = 6.28 mT; at its middle the field is 6.16 mT (98.1%), and at its ends 3.13 mT, about half.

In Ampère’s law, a loop round wire A (5 A out of the page) gives ∮B·dl = μ₀ × 5 = 6.28 µT m, wherever the loop sits.

Tips

• The field of a piece falls as 1/r², so near pieces matter most.
• Out of the page counts as positive current for a loop followed anticlockwise.
• Ampère’s law is easiest to use where the field is the same all round the loop (a long wire, a long solenoid).

FAQ

Why is the 24-piece loop a little low? The pieces are short straight chords, so the polygon is slightly smaller than the circle. More pieces would close the gap.

Do wires outside the Ampère loop matter? They change B at every point of the loop, but their contributions add up to zero round it.

Is this an official NCERT resource? No. fibub is independent; the page follows the topics in the syllabus in its own words.

Last updated 2026-11-09.