Lesson 5 of 12 · 7 min
Field on the axis of a circular loop
NCERT §4.5
Kabir winds 100 turns of wire into a flat coil of radius 5 cm and passes 2 A. How strong is the field at its centre, and how fast does it fade as he moves a probe along the axis?
The lesson in notes
In short
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.