Question

A toroid has a square cross section with the length of an edge equal to the radius of the inner surface. The ratio of the magnitude of the magnetic field at the inner surface to the magnitude of the field at the outer surface is :

a) 1/4
b) 1/2
c) 1
d) 2

Answer 1

The ratio of the magnetic field at the inner surface to the magnetic field at the outer surface of a toroid with a square cross-section and an edge equal to the inner radius is 1/2.

Step-by-step explanation:

The question revolves around the concept of the magnetic field in a toroid with a square cross section, the toroid being a type of inductor used in electronic circuits. The magnetic field (B) inside a toroid is given by Ampère's Law, which states that B = (μ_0 * N * I) / (2 * π * r), where μ_0 is the permeability of free space, N is the number of turns, I is the current passing through the wire, and r is the radial distance from the center of the toroid. Given this, the ratio of the magnetic field at the inner surface to the outer surface depends on the ratio of their radial distances. With the inner and outer radii designated as rin and rout, the ratio Bin/Bout is simply rout/rin because the other factors (μ_0, N, and I) are constant. If the edge of the square cross section is equal to the inner radius, the outer radius will be twice the inner radius, making the ratio 1/2 (or b) when considering a toroid with a square cross-section.