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A 37 cm long solenoid, 1.8 cm in diameter, is to produce a 0.50 T magnetic field at its center. If the maximum current is 4.4 A, how many tu
Question
A 37 cm long solenoid, 1.8 cm in diameter, is to produce a 0.50 T magnetic field at its center. If the maximum current is 4.4 A, how many turns must the solenoid have? Express your answer using two significant figures.
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Physics
4 years
2021-08-21T20:40:45+00:00
2021-08-21T20:40:45+00:00 2 Answers
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Answers ( )
Answer:
33,458.71 turns
Explanation:
Given: L = 37 cm = 0.37 m, B= 0.50 T, I = 4.4 A, n= number of turn per meter
μ₀ = Permeability of free space = 4 π × 10 ⁻⁷
Solution:
We have B = μ₀ × n × I
⇒ n = B/ (μ₀ × I)
n = 0.50 T / ( 4 π × 10 ⁻⁷ × 4.4 A)
n = 90,428.94 turn/m
No. of turn through 0.37 m long solenoid = 90,428.94 turn/m × 0.37
= 33,458.71 turns
Given Information:
Diameter of solenoid = d = 1.8 cm = 0.018 m
Length of solenoid = L = 37 cm = 0.37 m
Current = I = 4.4 A
Magnetic field = B = 0.50 T
Required Information:
Number of turns = N = ?
Answer:
Number of turns ≈ 33,498 or 33,458
Step-by-step explanation:
The magnetic field at the center of the solenoid is given by
B = μ₀NI/√
(L²+4r²)
N = B√
(L²+4r²)/μ₀I
Where L is the length and r is the radius of the solenoid, N is the number of turns and B is the magnetic field.
r = d/2 = 0.018/2 = 0.009 m
N = 0.50√
(0.37)²+(4*0.009²)/4πx10⁻⁷*4.4
N ≈ 33,498 Turns
Please note that we can also use a more simplified approximate model for this problem since the length of the solenoid is much greater than the radius of the solenoid
L = 0.37 >> r = 0.009
The approximate model is given by
B = μ₀NI/L
N = BL/μ₀I
N = 0.50*0.37/4πx10⁻⁷*4.4
N ≈ 33,458 Turns
As you can notice the results with the approximate model are very close to the exact model.