Inside and outside a thick wire.

B against r inside and outside a solid wire, hollow tube or thin wire, with the Amperian circle at the probe radius and the enclosed current shaded.

For a 5 mm solid wire, find both radii where B is half its surface value (2.5 mm and 10 mm)

Using μ₀I/2πr inside the wire.

• Take any open surface with a boundary loop. Ampere's law says the line integral of B round the boundary equals μ₀ times the net current through the surface: ∮B·dl = μ₀I.
• Sign convention: curl the right-hand fingers in the sense the loop is traversed; the thumb shows which way a current counts as positive.
• When a loop can be chosen so that B is tangential and constant along part of length L, and normal to the loop or zero elsewhere, the law reduces to BL = μ₀Iₑ, with Iₑ the enclosed current.
• Long straight wire: a circle of radius r round it gives B × 2πr = μ₀I, so B = μ₀I/2πr. The field is the same all round the circle (cylindrical symmetry), tangential to it, and falls as 1/r.
• The field lines round a wire are closed circles, unlike electric field lines, which start and end on charges. Right-hand rule for a straight wire: thumb along the current, fingers curl the way B goes.
• Example 4.7: for a thick wire of radius a with current spread uniformly, B = μ₀I/2πr outside (r > a) and B = (μ₀I/2πa²)r inside (r < a). B grows linearly inside, peaks at the surface, then falls as 1/r.
• Ampere's law carries the same physics as the Biot–Savart law, the way Gauss's law does for Coulomb's law. It holds for steady currents and any loop, but it only gives B easily when there is enough symmetry; it cannot give μ₀I/2R at the centre of a loop.

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