You have a 2m long wire which you will make into a thin coil with N loops to generate a magnetic field of 3mT when the current in the wire i

Question

You have a 2m long wire which you will make into a thin coil with N loops to generate a magnetic field of 3mT when the current in the wire is 1.2A. What is the radius of the coils and how many loops, N, are there

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RobertKer 6 months 2021-08-19T19:18:25+00:00 1 Answers 7 views 0

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    2021-08-19T19:19:27+00:00

    Answer:

    radius of the loop =  7.9 mm

    number of turns N ≅ 399 turns

    Explanation:

    length of wire L= 2 m

    field strength B = 3 mT = 0.003 T

    current I = 12 A

    recall that field strength B = μnI

    where n is the turn per unit length

    vacuum permeability μ  = 4\pi *10^{-7}  T-m/A = 1.256 x 10^-6 T-m/A

    imputing values, we have

    0.003 = 1.256 x 10^−6 x n x 12

    0.003 = 1.507 x 10^-5 x n

    n = 199.07 turns per unit length

    for a length of 2 m,

    number of loop N = 2 x 199.07 = 398.14 ≅ 399 turns

    since  there are approximately 399 turns formed by the 2 m length of wire, it means that each loop is formed by 2/399 = 0.005 m of the wire.

    this length is also equal to the circumference of each loop

    the circumference of each loop = 2\pi r

    0.005 = 2 x 3.142 x r

    r = 0.005/6.284 = 7.9*10^{-4} m = 0.0079 m = 7.9 mm

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