A metal ring 5.00 cm in diameter is placed between the north and south poles of large magnets with the plane of its area perpendicular to th

Question

A metal ring 5.00 cm in diameter is placed between the north and south poles of large magnets with the plane of its area perpendicular to the magnetic field. These magnets produce an initial uniform field of 1.12 T between them but are gradually pulled apart, causing this field to remain uniform but decrease steadily at 0.260 T/s .

(a) What is the magnitude of the electric field induced in the ring?

(b) In which direction (clockwise or counterclockwise) does the current flow as viewed by someone on the south pole of the magnet?

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niczorrrr 3 years 2021-07-26T12:13:08+00:00 1 Answers 18 views 0

Answers ( )

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    2021-07-26T12:14:50+00:00

    Answer:

    a) 6.49*10^-3 N/C

    b) counterclockwise

    Explanation:

    a) To find the electric field induced in the ring you use the following formula or Faraday’s law:

    \int Eds=E(2\pi r)=-\frac{d\Phi_B}{dt}\\\\E=-\frac{1}{2\pi r}\frac{d\Phi_B}{dt}

    \Phi_B=AB=\pi r^2 B

    E=-\frac{1}{2\pi r}A\frac{dB}{dt}

    E: electric field

    Ф: magnetic flux

    A: area of the ring = pi*(0.05m)^2=7.85*10^-3 m^2

    dB/dt: change in the magnetic flux = -0.260T/s

    By replacing all these values of the parameters with r=5.00cm, in order to calculate E in the ring, you obtain:

    E=-\frac{1}{2\pi (0.05m)}(7.85*10^{-3}m^2)(-0.260T/s)=6.49*10^{-3}N/C

    b) induced current generate a magnetic field that is opposite to the first magnetic field. By this reason the induced current is counterclockwise.

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