Answer :
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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