Your employer asks you to build a 24-cm long solenoid with an interior field of 4.6 mT. The specifications call for a single layer of wire,

Question

Your employer asks you to build a 24-cm long solenoid with an interior field of 4.6 mT. The specifications call for a single layer of wire, wound with the coils as close together as possible. You have two spools of wire available. Wire with a #18 gauge has a diameter of 1.02 mm and has a maximum current rating of 6 A. Wire with a #26 gauge is 0.41 mm in diameter and can carry up to 1 A.

a. Which wire should you use?

b. What current will you need?

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Khoii Minh 5 years 2021-07-27T11:31:17+00:00 1 Answers 368 views 0

Answers ( )

  1. Vodka
    0
    2021-07-27T11:33:07+00:00
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    Answer:

    Explanation:

    Formula for magnetic field in a solenoid

    B = μ₀ N/ L x I

    N is total number of turns , L is length of solenoid , I is current , B is magnetic field ,  μ₀ is constant equal to 4π x 10⁻⁷

    Putting the values for 18 gauge wire

    N = length of wire / diameter of wire

    N = 24 / .102

    = 235.3

    B = μ₀ N/ L x I

    4.6 x 10⁻³ = 4π x 10⁻⁷ x (235.3 / .24) x I

    I = 3.735 A

    Putting the values for 26 gauge wire

    N = length of wire / diameter of wire

    N = 24 / .041

    = 585.36

    B = μ₀ N/ L x I

    4.6 x 10⁻³ = 4π x 10⁻⁷ x (585.36 / .24) x I

    I = 1.5  A

    a ) So 18 gauge wire will be used

    b ) as current required in this case is

    3.735 A

    In the second case current required is 1.5 A which is beyond its tolerence limit.

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