Question:

A toroid with mean radius r0 and diameter 2a has N turns carrying current I. What is the magnetic field B inside the toroid?

μ₀NI/2πr₀

μ₀NI/2π(r₀+a)

Zero

μ₀NI/π(r₀+a)

Along path 2, inside the toroid, we apply Ampere's Law ∮B.dl = μ₀ienc. B × 2πr₀ = μ₀ × NI. Therefore, B = μ₀NI/(2πr₀).