5.0-A current. (a) Find the magnetic field at a point along the axis of the coil, 0.80 m from the center. (b) Along the axis, at what distan

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5.0-A current. (a) Find the magnetic field at a point along the axis of the coil, 0.80 m from the center. (b) Along the axis, at what distance from the center of the coil is the field magnitude 1 8 as great as it is at the center

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Bơ 4 years 2021-07-27T10:34:00+00:00 1 Answers 62 views 0

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  1. danthu
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    2021-07-27T10:35:45+00:00
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    Here is the full question

    A coil consisting of 100  circular loops with radius 0.6 m carries a 5.0-A current.

    (a) Find the magnetic field at a point along the axis of the coil, 0.80 m from the center.

    (b) Along the axis, at what distance from the center of the coil is the field magnitude 1/8 as great as it is at the center

    Answer:

    a) B = 0.000113 T

    b) The distance (x) from the center of the coil = 1.038 m

    Explanation:

    The formula for the magnetic field on the line of axis of the circular loop can be expressed as:

    B = \frac{\mu NIr^2}{2 ( r^2+x^2)^{3/2}}

    where;

    N = number of turns = 100

    \mu = 4 \pi * 10^{-7} \ N/m

    I = current flowing through the coil = 5.0- A

    r = radius of the circular loop = 0.6 m

    x = distance from the center of the coil

    So;

    B = \frac{4 \pi *10^{-7}N/m (100)(5A)(0.6\ m^2)}{2((0.6 \m )^2+(0.8m)^2)^{3/2}}

    B = 0.000113 T

    b)

    Also; to determine the distance (x) from the center of the coil; we have the following:

    We know that the magnetic field at the center of the coil can be expressed as:

    B = \frac{\mu_o NI}{2 r}

    Now; given that the field magnitude is 1/8 as great as it is at the center; Then ;

    B' = \frac{1}{8} B

    ∴

    \frac{\mu NIr^2}{2 ( r^2+x^2)^{3/2}} = \frac{1}{8} ( \frac{\mu_o NI}{2 r})

    \frac{r^2}{(r^2+x^2)^{3/2}}= (\frac{1}{8}) r

    8r^3 = (r^2+x^2)^{3/2}

    (8r^3)^2 = (r^2+x^2)^3

    (64r^6) = (r^2+x^2)^3

    (r^2+x^2) = \sqrt[3]{64r^6}

    (r^2+x^2) = 4r^2

    x^2 = 4r^2- r^2

    x^2 = 3 \ r^2

    x= \sqrt {3 r^2}

    x = 1.73 (r)

    x = 1.73 (0.6)

    x = 1.038 \ m

    Therefore, the distance from the center of the coil is  = 1.038 m

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