Consider a toroidal structure with a rectangular cross-section. If the toroid is defined by the surfaces r = 1cm and r = 4cm and the planes z = 0 cm and z = 2 cm, and the surface current density on the surface defined by r = 4cm is given by -60az A/m. (a) Specify the current densities on the surfaces at r = 1, z = 0, and z = 2. (b) Find the expression for H inside the toroid (i.e. in the region 1 < r < 4 cm and 0 < z < 2cm). I have the solution, I just don't really understand it. The current density, K, for r=1 was found to be 240 A/m. To find the field inside the toroid ampere's law is used [itex]\oint[/itex]H dl = 2∏r = ∫Kr=1dθ from θ=0 to 2∏. Giving the final answer of H=15/(2∏r) So in this question K is different at r=4 and r=1. Is it changing due to distance from the sources of the field, or is it uniform and I'm missing some simple math? Since K is different at different radii, how come we integrate only integrate with Kr=1 and not other any other K's? I mean it is changing and I'm assuming since the cross section of the toroid is a square loop, all four sides would contribute to the field inside the toroid, not just r=1 . Wouldn't I need to calculate for Kr=4 and Kz=0 =Kz=2 and sum them up? Hope that made sense, thanks in advance.