Por una espira de 0.5 m2 de área circula una corriente de 5 A. Calcula la densi- dad de flujo magnético B considerando que la espira conside

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Por una espira de 0.5 m2 de área circula una corriente de 5 A. Calcula la densi- dad de flujo magnético B considerando que la espira considera la permeabilidad del medio es la del aire

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Ngọc Khuê 3 years 2021-09-03T15:41:10+00:00 1 Answers 140 views 0

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  1. taiduc
    0
    2021-09-03T15:42:24+00:00
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    Answer:

    \beta=B=8.05\mu T

    Explanation:

    The density of the magnetic flux is given by the following formula:

    \beta=\frac{\Phi_B}{A}=\frac{ABcos\theta}{A}=Bcos\theta

    The normal vector A and the vector of the magnitude of the magnetic field are perpendicular, then, the angle is zero:

    The magnitude of the magnetic field is calculated by using the formula for B at a distance of x to a point in the plane of the loop:

    B=\frac{\mu_oIR^2}{2(x^2+R^2)^{3/2}}

    For x = 0 you have:

    B=\frac{\mu_oIR^2}{2R^3}=\frac{\mu_oI}{2R}

    R is the radius of the circular loop and its values is:

    R=\sqrt{\frac{A}{\pi}}=\sqrt{\frac{0.5m^2}{\pi}}=0.39m

    Then, you replace in the equation for B with mu_o = 4\pi*10^-7 T/A:

    B=\frac{(4\pi*10^{-7}T/A)(5A)}{2(0.39m)}=8.05*10^{-6}T=8.05\mu T

    and the density of the magnetic flux is

    \beta=B=8.05\mu T

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