A conducting wire formed in the shape of a right triangle with base b = 34 cm and height h = 77 cm and having resistance R = 2.2 Ω, rotates

Question

A conducting wire formed in the shape of a right triangle with base b = 34 cm and height h = 77 cm and having resistance R = 2.2 Ω, rotates uniformly around the y-axis in the direction indicated by the arrow (clockwise as viewed from above (loooking down in the negative y-direction)). The triangle makes one complete rotaion in time t = T = 1.6 seconds. A constant magnetic field B = 1.6 T pointing in the positive z-direction (out of the screen) exists in the region where the wire is rotating.1)What is ?, the angular frequency of rotation?2)What is Imax, the magnitude of the maximum induced current in the loop?3)At time t = 0, the wire is positioned as shown. What is the magnitude of the magnetic flux ?1 at time t = t1 = 0.45 s?4)What is I1, the induced current in the loop at time t = 0.45 s? I1 is defined to be positive if it flows in the negative y-direction in the segment of length h.

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Ngọc Khuê 4 years 2021-07-19T05:22:21+00:00 1 Answers 25 views 0

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  1. nguyetanh
    0
    2021-07-19T05:23:22+00:00
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    Answer:

    a) 3.92 rad/s

    b) 0.373 A

    d) 0.018 A

    Explanation:

    a) The angular frequency of rotation is given by:

    \omega=\frac{2\pi}{T}=\frac{2\pi}{(1.6s)}=3.92\frac{rad}{s}

    b) The maximum induced current in the loop is given by:

    I_{max}=\frac{emf_{max}}{R}=\frac{AB\omega}{R}

    R: resistance

    A: area of the triangle loop = bh/2 = (0.34m)(0.77m)/(2) = 0.1309m^2

    B: magnitude of the magnetic field

    I_{max}=\frac{(0.13m^2)(1.6T)(3.92rad/s)}{2.2\Omega}=0.373A

    d) For t = 0.45s you have:

    I(t)=\frac{ABcos(\omega t)}{R}\\\\I(0.45)=\frac{(0.13m^2)(1.6T)cos(3.92rad/s \ (0.45s))}{2.2\Omega}=-0.018A

    But I1 is defined to be positive if it flows in the negative y-direction.

    hence, I for t=0.45 s is 0.018A

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