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------------------------------------------------ An air-filled 39.1 μH solenoid has a length of 4.0 cm and a cross-sectional area of 0.60 cm². How many turns are in this solenoid?

A. 1200
B. 12
C. 120
D. 144
E. 21000

To find the number of turns in the solenoid, we can use the formula for the inductance of a solenoid:
L = μ₀ * (N^2 * A) / l

where L is the inductance (39.1 micro H), μ₀ is the permeability of free space (4π × 10^-7 Tm/A), N is the number of turns, A is the cross-sectional area (0.60 cm^2), and l is the length of the solenoid (4.0 cm). We need to solve for N.

Explanation:

First, let's convert the given values into the SI unit system:
L = 39.1 × 10^-6 H
A = 0.60 × 10^-4 m^2
l = 4.0 × 10^-2 m

Now, rearrange the formula to solve for N:
N^2 = (L * l) / (μ₀ * A)

Substitute the given values:
N^2 = (39.1 × 10^-6 * 4.0 × 10^-2) / (4π × 10^-7 * 0.60 × 10^-4)

Calculate the result:
N^2 ≈ 144

Take the square root to find the number of turns:
N ≈ √144 = 12

So, the correct answer is (b) 12 turns in the solenoid.

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