Which expression correctly gives the magnetic field inside an ideal solenoid with turns per unit length n and current I?

Which expression correctly gives the magnetic field inside an ideal solenoid with turns per unit length n and current I?

• A. μ0 I
• B. μ0 I n^2
• C. μ0 n
• D. μ0 n I

Answer: (D)

Ampere's law applied to a long, ideal solenoid tells us the magnetic field inside is uniform and proportional to the current and the turns per unit length. Consider a rectangular loop of length l inside the solenoid; the field is constant and parallel to the loop, so ∮ B·dl = B l. The current enclosed by this loop is the number of turns in length l times the current per turn, which is (n l)I. Ampere's law gives B l = μ0 (n l I), and canceling l yields B = μ0 n I. This matches the expected behavior: more turns per unit length or larger current increases the field inside. Outside, an ideal long solenoid has essentially zero field, reinforcing why the inside expression is clean. The units work out as μ0 (T·m/A) times n (1/m) times I (A), giving Tesla. So the correct expression is μ0 n I.

Ampere's law applied to a long, ideal solenoid tells us the magnetic field inside is uniform and proportional to the current and the turns per unit length. Consider a rectangular loop of length l inside the solenoid; the field is constant and parallel to the loop, so ∮ B·dl = B l. The current enclosed by this loop is the number of turns in length l times the current per turn, which is (n l)I. Ampere's law gives B l = μ0 (n l I), and canceling l yields B = μ0 n I. This matches the expected behavior: more turns per unit length or larger current increases the field inside. Outside, an ideal long solenoid has essentially zero field, reinforcing why the inside expression is clean. The units work out as μ0 (T·m/A) times n (1/m) times I (A), giving Tesla. So the correct expression is μ0 n I.