Ampere's Law — Thick ConductorHardQuestion 20233
Question
A long cylindrical conductor of inner radius a and outer radius b carries current I uniformly distributed through the annular cross-section. At a point r where a < r < b, B is:
- Aμ₀I/2πr
- Bμ₀I(r²−a²)/[2πr(b²−a²)]Correct
- Cμ₀I(r−a)/[2π(b−a)]
- DZero
Correct answer
μ₀I(r²−a²)/[2πr(b²−a²)]
Explanation
Current density J = I/[π(b²−a²)]. Current enclosed by circle of radius r: I_enc = J·π(r²−a²) = I(r²−a²)/(b²−a²). Ampere's law: B(2πr) = μ₀I_enc → B = μ₀I(r²−a²)/[2πr(b²−a²)].