Find the magnetic field of a toroidal coil using Ampere's Law.

We compute the magnetic field of a toroid by applying Ampere's Law. 🧠 Access full flipped physics courses with video lectures and examples at https://www.zakslabphysics.com/ 00:00 Introduction and key assumptions: we are given a toroidal coil with N turns carrying a current of I. We need to make a couple reasonably simplifying assumptions for the toroidal inductor based on our prior experience with the long solenoid. First, we recall that the field was zero outside a long solenoid, so we assume that when we bend the solenoid into the shape of a donut, the field will still be zero outside. Second, we recall that the field inside a solenoid is parallel to the solenoid axis, so now we assume that the field continues to follow the axis of the circular cross sections of the toroid, in other words the magnetic field of a toroid is circumferential! We choose a circumferential Amperian loop in order to line up with this circumferential field. 01:58 Determine the direction of magnetic field within a toroid: Next, we view a schematic representation of the toroid cross section (cut through the symmetry plane like a bagel). We apply the right hand rule to show that all wires give a counterclockwise field component within the toroid cross section, so we conclude the magnetic field is perfectly counterclockwise within the toroid. 04:07 Magnetic field of a toroidal coil using Ampere's Law: now that everything is set up, we can calculate the Ampere's law path integral on the left hand side. The magnetic field vector is tangent to the Amperian loop at every point, making it parallel to the path increments dl. Thus, the dot product is trivial. By symmetry, B is constant along the Amperian loop, allowing us to factor it out of the integral to obtain B(2pi*r) on the left hand side. On the right side of Ampere's Law, we need to compute the current enclosed by the Amperian loop, but that's just the number of turns multiplied by the current I, giving us mu_0*NI for the right hand side. We quickly solve for B to find the magnetic field inside a toroid: mu_0*N*I/(2pi*r). It is interesting to note the contrast with a solenoid field: the solenoid field was uniform throughout the interior, but the toroid magnetic field drops off like 1/r, where r is the distance to the symmetry axis of the toroid. #physics #magnetism #amperecircuitallaw