Final answer:
To calculate the number of turns in the solenoid, the formula B = μ0 * (N/L) * I is used. After inserting the given values (B = 4.0 T, I = 3.0 A, and L = 0.10 m), the result is approximately 100,000 turns for the solenoid.
Step-by-step explanation:
The student is asking how many turns make up a solenoid that generates a magnetic field of 4.0 T when a current of 3.0 A flows through it, given that the solenoid has a radius of 0.020 m and is 0.10 m long. To calculate the number of turns (N) in the solenoid, we can use the formula for the magnetic field (B) inside a long solenoid:
B = μ₀ * (N/L) * I
Where:
- μ₀ is the permeability of free space (4π x 10^-7 T·m/A)
- N is the number of turns
- L is the length of the solenoid
- I is the current
Given:
- B = 4.0 T
- I = 3.0 A
- L = 0.10 m
Rearranging the formula to solve for N:
N = (B * L) / (μ₀ * I)
Plugging in the known values:
N = (4.0 T * 0.10 m) / (4π x 10^-7 T·m/A * 3.0 A) ≈ 10^5 turns
Therefore, the solenoid is comprised of approximately 100,000 turns.