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1 November, 21:45

If the frequency of the voltage signal impressed on the parallel plate capacitor is doubled from f to 2f, the amplitude of the magnetic field Bo at r = a at the mid-plane of the gap between the plates of this capacitor will: Increase by a factor of two. Remain the same. Decrease by a factor of four. Decrease by a factor of two. Increase by a factor of four.

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  1. 1 November, 23:16
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    Increase by a factor of four

    Explanation:

    Given data

    capacitor = f to 2f

    r = a

    to find out

    capacitor will be

    solution

    first we calculate the current

    displacement current will be = ɛ * electric flux

    displacement current will be = ɛ * electric flux

    electric flex = A * electric field

    electric field = voltage / d

    electric field = V (o) sin ωt / d

    so electric flex

    electric flex = A * V (o) sin ωt / d

    and displacement current will be

    displacement current = ɛ * A * V (o) sin ωt / d

    so the magnetic filed will be here

    magnetic filed = μ * current + μ * ɛ d∅/dt

    so we can say that frequency will be double when magnetic field increase

    so Increase by a factor of four
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