Physics · Moving Charges And Magnetism · NEET
Both are correct, and they are the same equation. The full form is B = mu0 N I / (2 pi r), where N is the TOTAL number of turns. Since the turns per unit length is n = N / (2 pi r) (the circumference of the circle of radius r), substituting gives B = mu0 n I. Use B = mu0 N I / (2 pi r) when the problem gives you total turns N and radius r; use B = mu0 n I when it directly gives turns per metre n.
Apply Ampere's law to a circular loop drawn OUTSIDE the toroid (radius bigger than the outer edge). That loop encloses N turns carrying current I one way and the same N turns of the return path effectively, so the net enclosed current is zero. Since the enclosed current is zero, the line integral of B is zero, so B = 0 outside. The same is true for a loop through the central hole: it encloses no current, so B = 0 there too.
N (capital) is the TOTAL number of turns wound on the whole toroid, a plain number like 1000. n (small) is turns per unit length, n = N / (2 pi r), measured in turns per metre. Plugging N into the n-formula (or the reverse) is the most common toroid mistake. Check units: B = mu0 N I / (2 pi r) needs r in the denominator, but B = mu0 n I does not.
In a straight solenoid the field is uniform along the axis, so B = mu0 n I has no r term. In a toroid the same turns are packed around circles of different radii, so the turns-per-length is larger on the inner edge and smaller on the outer edge. That makes B larger near the inner wall and smaller near the outer wall, so B varies as 1/r across the tube. For NEET, if only the mean radius is given, use that r.
Choose a CIRCLE that is concentric with the toroid and passes through the point where you want B, lying inside the tube. By symmetry B is constant in magnitude and tangential along this circle, so the integral of B.dl becomes B times (2 pi r). Set that equal to mu0 times enclosed current (N I) to get B = mu0 N I / (2 pi r).
Try the real previous-year questions from this chapter — each with the answer and a full solution.
No. It varies as 1/r across the tube, so it is stronger near the inner wall and weaker near the outer wall. It is only approximately uniform when the tube is thin compared with the mean radius, and NEET usually treats it that way using the mean radius r.
Zero. An Amperian circle drawn through the central empty hole encloses no current, so B = 0 there, just like outside the toroid.
Yes. If the toroid has a magnetic core of permeability mu, replace mu0 with mu, so B = mu N I / (2 pi r). Iron cores greatly increase the field, which is why transformers and inductors use toroidal cores.
A solenoid is a straight coil; a toroid is a solenoid bent into a closed ring. The solenoid field B = mu0 n I is uniform and has small end leakage, while the toroid field B = mu0 N I / (2 pi r) is fully confined inside with no external field.
For the field at a specific point, use the radius of the circle passing through that point. If the problem only gives inner and outer radii, use the mean radius r = (r_inner + r_outer)/2 for an average field, unless it asks for the field at a stated radius.
