A circular current‑carrying loop lies so that the plane of the loop is perpendicular to a constant magnetic field of strength B. Suppose tha

Question

A circular current‑carrying loop lies so that the plane of the loop is perpendicular to a constant magnetic field of strength B. Suppose that the radius R of the loop could be made to increase with time t so that R = at, where a is a constant. What is the magnitude of the emf that would be generated around the loop as a function of t ?

in progress 0
Dâu 4 years 2021-07-20T13:12:19+00:00 1 Answers 34 views 0

Answers ( )

  1. Nho
    0
    2021-07-20T13:13:35+00:00
    Reply

    Answer:

    The magnitude of the emf  is  e = 2 \pi B a^2 t

    Explanation:

    Let  e be the magnitude of the emf that would be generated

        Since the loop is carrying current then there will be magnetic flux flowing through the loop area, let denote this magnetic flux as \O_B and can be mathematically represented as

                     \O_B = BAcos \theta

    Where B is the magnetic field

               A area of the loop

    The change magnetic flux with time is mathematically represented as

                   \frac{\Delta \O_B}{\Delta t} = B\frac{\Delta A}{\Delta t} cos(\theta)

      From this equation we see that the change of magnetic flux with time as changes the area with time

         Generally Area is mathematically represented as

                                 A = \pi R^2

    From the question we are told that the radius  is

                            R = at

    substituting this into the equation for A

                             A = \pi (at)^2

                              A = a^2 \pi t^2

    Now the change of area with would be mathematically evaluated as

                         \frac{\Delta A}{\Delta t} = \frac{d}{dt} a^2 (\pi t^2)

                                = 2\pi t a^2

    Given that \theta =0 since the plane of the  loop is  perpendicular to the constant magnetic field of strength B

       Now e can be represented mathematically as

                     e =N \frac{d \O_B}{dt} = N B \frac{dA}{dt}

    Where N is the number of turns  = 1  that is a loop

                     e = B \frac{dA}{dt}

    substituting for change in area with time

               e = 2 \pi B a^2 t

                   

                       

                       

Leave an answer

Browse

Giải phương trình 1 ẩn: x + 2 - 2(x + 1) = -x . Hỏi x = ? ( )