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16 November, 09:35

A circular coil (800 turns, radius = 0.063 m) is rotating in a uniform magnetic field. At t = 0 s, the normal to the coil is perpendicular to the magnetic field. At t = 0.016 s, the normal makes an angle of 45o with the field because the coil has made one-eighth of a revolution. An average emf of magnitude 0.042 V is induced in the coil. Find the magnitude of the magnetic field at the location of the coil.

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  1. 16 November, 11:07
    0
    0.000099T

    Explanation:

    For a circular coil the flux through a single turn changes by:

    ∆ = BAcos45° - BAcos90°

    = BA cos 45°

    During the interval of ∆t = 0.0165s

    For N turn, Faraday's law gives the magnitude of emf as follows:

    /E / = / - N BAcos45° / : ∆t

    Since the loops are circular the area A of each loop is equal to πr square

    B = / E / ∆t : Nπr square cos45°

    = 0.042V * 0.0165 / 800 * 3.142 * 0.063^2 cos 45°

    = 0.000693/7

    = 0.000099T

    = 9.9*10^-5T
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