Inside a long solenoid with N turns and length l carrying current I, the magnetic field is?

Inside a long solenoid with N turns and length l carrying current I, the magnetic field is?

• A. B = μ0 N I / l
• B. B = μ0 n I with n = N/l
• C. B = μ0 I /(2π r)
• D. B = μ0 N I

Answer: (B)

Inside a long solenoid, the magnetic field is nearly uniform and runs along the axis. The strength grows with both the current and how densely the windings are packed, which is captured by B = μ0 n I, where n is the number of turns per unit length. If the solenoid has N turns over a length l, then n = N/l, so B = μ0 (N/l) I. That is the same as B = μ0 N I / l, just written with the turn density n. The key idea is that the field scales with how many turns there are per unit length and with the current, not with the total number of turns alone. The other expressions don’t fit because they either describe a different geometry (a long straight wire, where B ∝ I/(2πr)) or neglect the length of the solenoid and the per-length turn density, which are essential for the correct magnitude inside a solenoid.

Inside a long solenoid, the magnetic field is nearly uniform and runs along the axis. The strength grows with both the current and how densely the windings are packed, which is captured by B = μ0 n I, where n is the number of turns per unit length. If the solenoid has N turns over a length l, then n = N/l, so B = μ0 (N/l) I. That is the same as B = μ0 N I / l, just written with the turn density n. The key idea is that the field scales with how many turns there are per unit length and with the current, not with the total number of turns alone.

The other expressions don’t fit because they either describe a different geometry (a long straight wire, where B ∝ I/(2πr)) or neglect the length of the solenoid and the per-length turn density, which are essential for the correct magnitude inside a solenoid.