Physics · Moving Charges And Magnetism · NEET
An Amperian loop is an imaginary closed path you draw to apply the law. You do not need a real wire on it. Choose it so B is either along the loop and constant, or perpendicular to it, or zero. For a long straight wire, pick a circle around the wire so B is tangential and same everywhere: then ∮ B·dl = B(2πr). Good symmetry makes the integral simple; a bad loop makes the integral impossible.
The law is always TRUE, but it only lets you FIND B easily when there is high symmetry (long straight wire, thick wire, solenoid, toroid). For the magnetic field at the centre of a single circular loop, no simple Amperian loop exists where B is constant and tangential, so you must use Biot-Savart law instead. NCERT states this directly.
Only the current that actually passes through the surface bounded by your loop — the ENCLOSED current I(e). If a wire is outside your Amperian loop, it adds nothing to ∮ B·dl. Inside a thick wire of radius a, for r < a only a fraction (r²/a²) of the current is enclosed, which is why B rises linearly with r inside.
For an ideal long solenoid the field outside is taken as zero and the field inside is uniform and parallel to the axis. When you draw a rectangular Amperian loop, only the side inside the solenoid contributes B·L; the outside side gives 0 and the two short sides are perpendicular to B (contribute 0). So ∮ B·dl = B·L = μ₀(nL)I, giving B = μ₀ n I.
No — they contain the same physics for steady currents. NCERT says 'Ampere's law is to Biot-Savart law what Gauss's law is to Coulomb's law.' Biot-Savart adds up field from tiny current elements; Ampere's law relates the field on a boundary to the current through it. Use Ampere's law when there is symmetry, Biot-Savart when there is not.
A long solenoid of radius 1 mm has 100 turns per mm. If 1 A current flows in the solenoid, the magnetic field strength at the centre of the solenoid is:
A long solenoid of 50 cm length having 100 turns carries a current 2.5 A. The magnetic field at the centre of the solenoid is (μ₀ = 4π × 10⁻⁷ T m A⁻¹):
Try the real previous-year questions from this chapter — each with the answer and a full solution.
∮ B·dl = μ₀ I(enclosed). The closed-loop line integral of the magnetic field equals μ₀ times the net current passing through the surface bounded by the loop.
μ₀ is the permeability of free space, μ₀ = 4π × 10⁻⁷ T·m/A (also written 4π × 10⁻⁷ H/m). Keep it in this form for fast NEET arithmetic with π.
1) Choose an Amperian loop matching the current's symmetry so B is tangential and constant (or zero). 2) Write ∮ B·dl = B × (loop length where B is along it). 3) Set it equal to μ₀ × enclosed current and solve for B.
B × 2πr = μ₀ I, so B = μ₀ I / (2πr). The field forms concentric circles around the wire and decreases as 1/r.
There is no closed path around that point on which B is both constant in magnitude and tangential, so the integral cannot be simplified. Use Biot-Savart law, which gives B = μ₀ I / 2R at the centre.
The law holds for ANY loop, but only a well-chosen symmetric loop lets you pull B out of the integral and solve. A poorly chosen loop still obeys the law but cannot give B.
