Field at the centre of an arc.

Field at the centre of a circular arc of chosen angle (straight leads add nothing), and B along the axis of a full loop.

Make a 2 cm semicircle carry enough current for 1.9 × 10⁻⁴ T at its centre (12 A)

Using μ₀I/2πr at a loop's centre.

• For a loop of radius R carrying I, every element is perpendicular to the line joining it to a point on the axis, so |dl × r| = r dl with r² = x² + R².
• Each element's field has a part along the axis and a part across it. The cross parts from opposite elements cancel, so only the axial parts add.
• Summing over the whole loop (total length 2πR) gives B = μ₀IR²/[2(x² + R²)^(3/2)] along the axis, at a distance x from the centre.
• At the centre (x = 0) this becomes B = μ₀I/2R. For N closely wound turns, multiply by N: B = μ₀NI/2R.
• The field lines of a loop are closed curves. Right-hand thumb rule for a loop: curl the fingers along the current, and the thumb points along the field through the loop. One face acts like a north pole, the other like a south pole.
• Example 4.5: straight leads make no field at a centre in line with them (dl × r = 0). A semicircle gives half a full loop's field, μ₀I/4R; for 12 A and R = 2.0 cm that is 1.9 × 10⁻⁴ T. Bending the arc the other way reverses the field but keeps its size.
• Example 4.6: a 100-turn coil of radius 10 cm with 1 A has B = μ₀NI/2R = 6.28 × 10⁻⁴ T at its centre.

Teaching this idea? Put the simulation on your own page or LMS with this code.