A circular loop in the plane of a paper lies in a 0.45 T magnetic field pointing into the paper. The loop’s diameter changes from 17.0 cm to

Question

A circular loop in the plane of a paper lies in a 0.45 T magnetic field pointing into the paper. The loop’s diameter changes from 17.0 cm to 6.0 cm in 0.53 s.
A) Determine the direction of the induced current.
B) Determine the magnitude of the average induced emf.
C) If the coil resistance is 2.5 Ω, what is the average induced current?

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Calantha 4 years 2021-08-08T12:09:37+00:00 1 Answers 29 views 0

Answers ( )

  1. Eirian
    0
    2021-08-08T12:10:44+00:00
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    Answer:

    (A). The direction of the induced current will be clockwise.

    (B). The magnitude of the average induced emf 16.87 mV.

    (C). The induced current is 6.75 mA.

    Explanation:

    Given that,

    Magnetic field = 0.45 T

    The loop’s diameter changes from 17.0 cm to 6.0 cm .

    Time = 0.53 sec

    (A). We need to find the direction of the induced current.

    Using Lenz law

    If the direction of magnetic field shows into the paper then the direction of the induced current will be clockwise.

    (B). We need to calculate the magnetic flux

    Using formula of flux

    \phi_{1}=BA\cos\theta

    Put the value into the formula

    \phi_{1}=0.45\times(\pi\times(8.5\times10^{-2})^2)\cos0

    \phi_{1}=0.01021\ Wb

    We need to calculate the magnetic flux

    Using formula of flux

    \phi_{2}=BA\cos\theta

    Put the value into the formula

    \phi_{2}=0.45\times(\pi\times(3\times10^{-2})^2)\cos0

    \phi_{2}=0.00127\ Wb

    We need to calculate the magnitude of the average induced emf

    Using formula of emf

    \epsilon=-N(\dfrac{\Delta \phi}{\Delta t})

    Put the value into t5he formula

    \epsilon=-1\times(\dfrac{0.00127-0.01021}{0.53})

    \epsilon=0.016867\ V

    \epsilon=16.87\ mV

    (C). If the coil resistance is 2.5 Ω.

    We need to calculate the induced current

    Using formula of current

    I=\dfrac{\epsilon}{R}

    Put the value into the formula

    I=\dfrac{0.016867}{2.5}

    I=0.00675\ A

    I=6.75\ mA

    Hence, (A). The direction of the induced current will be clockwise.

    (B). The magnitude of the average induced emf 16.87 mV.

    (C). The induced current is 6.75 mA.

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