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worty [1.4K]
3 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]3 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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The core of a 400 Hz aircraft transformer has a net cross-sectional area of 13 cm2. The maximum flux density is 0.9 T, and there
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Answer:

32.76 Volt

Explanation:

frequency, f = 400 Hz

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Maximum flux density, B = 0.9 tesla

Number of turns in secondary coil, N = 70

Let the maximum induced voltage is e.

According to the Faraday's law of electromagnetic induction, the induced emf is equal to the rate of change of magnetic flux.

e = dФ/dt

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Time is defined as the reciprocal of frequency.

So, e = N B A f

e = 70 x 0.9 x 13 x 10^-4 x 400

e = 32.76 volt

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Answer:

35

Explanation:

We are given that

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Final voltage, V_2=5 V

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We know that

\frac{N_s}{N_p}=\frac{V_2}{V_1}

Using the formula

\frac{N_s}{840}=\frac{5}{120}

N_s=\frac{5}{120}\times 840=35

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