The rotating loop in an AC generator is a square 10.0 cm on each side. It is rotated at 60.0 Hz in a uniform field of 0.800 TO. Calculate (a

Question

The rotating loop in an AC generator is a square 10.0 cm on each side. It is rotated at 60.0 Hz in a uniform field of 0.800 TO. Calculate (a) the flux through the loop as a function of time, (b) the emf induced in the loop, (c) the current induced in the loop for a loop resistance of 1.00 V, (d) the power delivered to the loop, and (e) the torque that must be exerted to rotate the loop.

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Khang Minh 4 years 2021-08-27T07:46:56+00:00 1 Answers 57 views 0

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  1. thienhuong
    0
    2021-08-27T07:48:49+00:00
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    Answer:

    Explanation:

    Given a square side loop of length 10cm

    L=10cm=0.1m

    Then, Area=L²

    Area=0.1²

    Area=0.01m²

    Given that, frequency=60Hz

    And magnetic field B=0.8T

    a. Flux Φ

    Flux is given as

    Φ=BA Sin(wt)

    w=2πf

    Φ=BA Sin(2πft)

    Φ=0.8×0.01 Sin(2×π×60t)

    Φ=0.008Sin(120πt) Weber

    b. EMF in loop

    Emf is given as

    EMF= -N dΦ/dt

    Where N is number of turns

    Φ=0.008Sin(120πt)

    dΦ/dt= 0.008×120Cos(120πt)

    dΦ/dt= 0.96Cos(120πt)

    Emf=-NdΦ/dt

    Emf=-0.96NCos(120πt). Volts

    c. Current induced for a resistance of 1ohms

    From ohms law, V=iR

    Therefore, Emf=iR

    i=EMF/R

    i=-0.96NCos(120πt) / 1

    i=-0.96NCos(120πt) Ampere

    d. Power delivered to the loop

    Power is given as

    P=IV

    P=-0.96NCos(120πt)•-0.96NCos(120πt)

    P=0.92N²Cos²(120πt) Watt

    e. Torque

    Torque is given as

    τ=iL²B

    τ=-0.96NCos(120πt)•0.1²×0.8

    τ=-0.00768NCos(120πt) Nm

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