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.

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 = ?  

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.

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