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worty [1.4K]
4 years ago
8

The magnetic field strength at the north pole of a 2.0-cm-diameter, 8-cm-long Alnico magnet is 0.10 T. To produce the same field

with a solenoid of the same size, carrying a current of 2.1 A , how many turns of wire would you need?
Physics
1 answer:
frosja888 [35]4 years ago
4 0

Answer:

N = 3032  turns

Explanation:

The magnetic field produced by a solenoid is described by

        B = μ₀ n I

Where is the permittivity in a vacuum with a value of 4π 10⁻⁷ N /A²,  n  is the turn density and I the current

Let's apply this equation to the problem, the turn density is the number of turns per unit length, in this case it is the same magnet length

       L = 8 cm = 0.08 m

Let's calculate

      B = μ₀ N/L   I

      N = B L / μ₀ I

      N = 0.10 0.08 / (4π 10⁻⁷  2.1)

      N = 3,032 103 turns

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Which describes a force acting on an object?
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<u>Answer</u>:

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<u>Explanation</u>:

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N_A+N_B = 9564.75 -------------(1)

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Solving the equation (1) and(2)

N_A + N_B = 9564.75

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F_B = 975a_G

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9564.75(2.20 -1.82) -F_B(0.55)=0

\frac{3634.605}{0.55}=F_B

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a_G_2 = 6.78m/s^2

Choosing the critical case

a_G = min(a_G_1 ,a_G_2)

a_G = min(7.848, 6.78)

a_G= 6.78 m/s^2

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