A 15.0 m \long copper wire, 2.20 mm in diameter including insulation, is tightly wrapped in a single layer with adjacent coils touching, to

Question

A 15.0 m \long copper wire, 2.20 mm in diameter including insulation, is tightly wrapped in a single layer with adjacent coils touching, to form a solenoid of diameter 2.10 cm (outer edge).a) What is the length of the solenoid?
b) What is the field at the center when the current in the wire is 16.7 A ?

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Thiên Thanh 2 months 2021-07-28T10:45:56+00:00 1 Answers 7 views 0

Answers ( )

    0
    2021-07-28T10:47:46+00:00

    a) 0.5 m

    b) 4.77\cdot 10^{-3} T

    Explanation:

    a)

    At the beginning, the length of the copper wire is:

    L = 15.0 m

    and the diameter (so, the thickness of each adjacent circle) is

    t=2.20 mm = 0.0022 m

    While the diameter of one circle of the solenoid is

    d=2.10 cm = 0.021 m

    So the perimeter of one circle is

    p=\pi d=\pi (0.021)=0.0659 m

    So the number of complete circles in the solenoid is

    n=\frac{L}{p}=\frac{15.0}{0.0659}=227.6

    The tickness of one circle is t, so the total length of the solenoid will be:

    L'=nt=(227.6)(0.0022)=0.5 m

    b)

    The magnetic field at the center of a solenoid is given by

    B=\mu_0 n I

    where

    \mu_0 is the vacuum permeability

    n is the number of turns of the solenoid

    I is the current in the solenoid

    Here we have:

    \mu_0 =4\pi \cdot 10^{-7}H/m is the vacuum permeability

    n=227.6 is the number of turns in the solenoid (calculated in part a)

    I = 16.7 A is the current in the solenoid

    Substituting, we find:

    B=(4\pi \cdot 10^{-7})(227.6)(16.7)=4.77\cdot 10^{-3} T

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