In a physics laboratory experiment, a coil with 250 turns enclosing an area of 14 cm2 is rotated in a time interval of 0.030 s from a positi

Question

In a physics laboratory experiment, a coil with 250 turns enclosing an area of 14 cm2 is rotated in a time interval of 0.030 s from a position where its plane is perpendicular to the earth’s magnetic field to a position where its plane is parallel to the field. The earth’s magnetic field at the lab location is 5.0×10^−5 T.Required:a. What is the total magnetic flux through the coil before it is rotated? After it is rotated? b. What is the average emf induced in the coil?

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Nem 3 years 2021-08-11T12:56:55+00:00 1 Answers 215 views 1

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  1. thuhuong
    1
    2021-08-11T12:58:12+00:00
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    Explanation:

    Consider a loop of wire, which has an area of A=14 \mathrm{cm}^{2} and N=250 turns, it is initially placed perpendicularly in the earth magnetic field. Then it is rotated from this position to a position where its plane is parallel to the field as shown in the following figure in \Delta t=0.030 s. Given that the earth’s magnetic field at the position of the loop is B=5.0 \times 10^{-5} \mathrm{T}, the flux through the loop before it is rotated is,

    \Phi_{B, i} &=B A \cos \left(\phi_{i}\right)=B A \cos \left(0^{\circ}\right

    =\left(5.0 \times 10^{-5} \mathrm{T}\right)\left(14 \times 10^{-4} \mathrm{m}^{2}\right)(1)

    =7.0 \times 10^{-8} \mathrm{Wb}

    \quad\left[\Phi_{B, i}=7.0 \times 10^{-8} \mathrm{Wb}\right

    after it is rotated, the angle between the area and the magnetic field is \phi=90^{\circ} thus,

    \Phi_{B, f}=B A \cos \left(\phi_{f}\right)=B A \cos \left(90^{\circ}\right)=0

    \qquad \Phi_{B, f}=0

    (b) The average magnitude of the emf induced in the coil equals the change in the flux divided by the time of this change, and multiplied by the number of turns, that is,

    {\left|\mathcal{E}_{\mathrm{av}}\right|=N\left|\frac{\Phi_{B, f}-\Phi_{B, i}}{\Delta t}\right|}{=} & \frac{1.40 \times 10^{-5} \mathrm{Wb}}{0.030 \mathrm{s}}

    & 3.6 \times 10^{-4} \mathrm{V}=0.36 \mathrm{mV}

    \mathbb{E}=0.36 \mathrm{mV}

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