The magnetic field inside a long solenoid having n turns per unit length and carrying current I is:
Aμ₀ n I / 2
Bμ₀ n I
Cμ₀ n I / (2π)
Dμ₀ I / (2πn)
Answer & Solution
Correct answer: B. μ₀ n I
1. Apply Ampère's law to a rectangular loop of side $h$ with one side inside the solenoid.
2. Only the inside side contributes: left side gives $Bh$.
3. Enclosed current is $I_e = I(nh)$, so $Bh = \mu_0 I(nh)$.
4. Cancel $h$: $B = \mu_0 n I$.
_Source: syllabus Physics Ch 4 "Moving Charges and Magnetism", p.153_
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