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2 September, 15:55

A circular coil (950 turns, radius = 0.060 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 = 0.010 s, the normal makes an angle of ϕ = 45° with the field because the coil has made one-eighth of a revolution. An average emf of magnitude 0.065 V is induced in the coil. Find the magnitude of the magnetic field at the location of the coil.

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  1. 2 September, 18:24
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    B = 8.55 * 10⁻⁵ T

    Explanation:

    Data Provided:

    Number of turns, N = 950 turns,

    Radius of coil, r = 0.060 m

    Now,

    Magnetic flux (φ) is given as:

    φ = BAcosθ

    where,

    B is the magnetic field

    A is the area

    Now,

    At t = 0 s, the normal to the coil is perpendicular to the magnetic field, i. e θ = 90°

    Thus,

    φ = BAcos90° = 0.

    Also,

    At t = 0.010 s, the normal makes an angle θ = 45°

    Thus,

    φ = NBAcos (45°),

    A = π * r²

    Average emf induced in the coil, E = 0.065 V.

    Thus,

    E = NBAcos (45°) / t

    or

    E = NB*πr²*cos (45°) / t

    or

    B = (E * t) / (NB*πr²*cos45°)

    on substituting the values

    we get,

    B = (0.065 V * 0.010 s) / (950 * π (0.06) ² * cos 45°)

    or

    B = 8.55 * 10⁻⁵ T
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