The magnetic field inside an ideal long solenoid with n turns per unit length carrying current I is
Aμ₀nI/2
Bμ₀I/(2πn)
C2πμ₀nI
Dμ₀nI
Answer & Solution
Correct answer: D. μ₀nI
1. Apply Ampere's law along a rectangular loop with one side inside and one outside.
2. The field outside is negligible; field inside is uniform and parallel to the axis.
3. Closed integral ∮B·dl = B·L = μ₀ · (nL · I).
4. Therefore B = μ₀nI inside an ideal long solenoid.
_Source: syllabus Physics Ch 4 "Moving Charges and Magnetism", §4.8.1_
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