A loop of wire is at the edge of a region of space containing a uniform magnetic field B⃗ . The plane of the loop is perpendicular to the ma

Question

A loop of wire is at the edge of a region of space containing a uniform magnetic field B⃗ . The plane of the loop is perpendicular to the magnetic field. Now the loop is pulled out of this region in such a way that the area A of the coil inside the magnetic field region is decreasing at the constant rate c. That is, dAdt=−c, with c>0.

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RI SƠ 5 years 2021-08-11T23:44:00+00:00 1 Answers 312 views 1

Answers ( )

  1. binhan
    1
    2021-08-11T23:45:25+00:00
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    QUESTION:

    Part A

    The induced emf in the loop is measured to be V. What is the magnitude B of the magnetic field that the loop was in?

    Part B

    For the case of a square loop of side length L being pulled out of the magnetic field with constant speed v (see the figure), what is the rate of change of area c = -\dfrac{dA}{dt}?

    Answer:

    Part A: B = -\dfrac{V}{c}

    Part B: c=-Lv

    Explanation:

    Part A:

    Faraday’s law says that the induced voltage is equal to

    V =-N \dfrac{d\Phi_B}{dt},

    which in our case(because we have only one loop) becomes

    V =- \dfrac{d (BA)}{dt},

    and since the magnetic field is uniform (not changing),

    V =-B \dfrac{dA}{dt}.

    Now, we know that \dfrac{dA}{dt} =c;

    therefore,

    V =-B c

    which gives us

    \boxed{B = -\dfrac{V}{c} }

    Part B:

    The area of the loop can be written as

    A = Lx,

    where x is the instantaneous length of the side along which the loop is moving.

    Taking the derivative of both sides we get:

    \dfrac{dA}{dt} = -L\dfrac{dx}{dt},

    and since v =\dfrac{dx}{dt} we have

    c = \dfrac{dA}{dt} = -Lv

    \boxed{c=-Lv}

    where the negative sign indicates that the area is decreasing.

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