alleryThe magnetic field inside a toroidal solenoid with N total turns, mean radius r, carrying current I is
Aμ₀NI
Bμ₀NI/(2πr)
C2πr·μ₀NI
Dμ₀NI/r
Answer & Solution
Correct answer: B. μ₀NI/(2πr)
1. Use Ampere's law along a circular path of radius r inside the toroid.
2. By symmetry, B is tangential: ∮B·dl = B·(2πr).
3. The enclosed current is NI.
4. Therefore B = μ₀NI/(2πr).
_Source: syllabus Physics Ch 4 "Moving Charges and Magnetism", §4.8.2_
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