A circular loop of flexible iron wire has an initial circumference of 165.0cm, but its circumference is decreasing at a constant rate of 12.

Question

A circular loop of flexible iron wire has an initial circumference of 165.0cm, but its circumference is decreasing at a constant rate of 12.0cm/s due to a tangential pull on the wire. The loop is in a constant, uniform magnetic field oriented perpendicular to the plane of the loop of magnitude 0.500T.

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Hưng Khoa 4 years 2021-08-16T01:34:01+00:00 2 Answers 272 views 0

Answers ( )

  1. thuthao
    0
    2021-08-16T01:35:01+00:00
    Reply

    Answer:

    Explanation:

    initial circumference, C = 165 cm = 1.65 m

    rate of change of circumference, dC/dt = 12 cm /s = 0.12 m/s

    magnetic field, B = 0.5 T

    According to the Faraday’s law of electromagnetic induction

    e = dФ/dt

    where, Ф is the magnetic flux

    Ф = B A

    where, A is the area of the coil

    e=\frac{d}{dt}(BA)

    e=B\frac{dA}{dt}

    e=B\frac{d(\pi r^{2})}{dt}

    e=2\pi r\times B\frac{dr}{dt}     … (1)

    C = 2πr

    dC/dt = 2π dr/dt

    Put in equation (1)

    e=C \times B\times \frac{1}{2\pi }\times \frac{dC}{dt}

    e = (1.65 x 0.5 x 0.12) / (2 x 3.14)

    e = 0.016 V

  2. Delwyn
    0
    2021-08-16T01:35:27+00:00
    Reply

    Answer:

    0.005 V

    Explanation:

    We are given that

    Initial circumference of circular loop=C=165 cm

    Rate of circumference,\frac{dC}{dt}=12 cm/s

    Magnetic field,B=0.5 T

    We have to find the induced emf at time t=9 s

    We know that induced amf,E=\frac{Bd(A)}{dt}

    Area of circular coil,A=\pi r^2

    E=B\frac{d(\pi r^2)}{dt}=B(2\pi r)\frac{dr}{dt}

    Circumference of circular coil,C=2\pi r

    165=2\pi r

    r=\frac{165}{2\pi}

    \frac{dr}{dt}=\frac{1}{2\pi}\frac{dC}{dt}=\frac{1}{2\pi}\times (12)=\frac{6}{\pi} cm/s=\frac{6\times 10^{-2}}{\pi} m/s

    Radius of coil at time t=9 s

    r=\frac{165}{2\pi}-(\frac{6}{\pi}\times 9)=9.08 cm=9.08\times 10^{-2} m

    1 m=100 cm

    E=-0.5(2\pi\times 9.08\times 10^{-2})\times \frac{6\times 10^{-2}}{\pi}=-0.005 V

    Magnitude of induced emf=0.005 V

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