The magnitude of the magnetic flux through the surface of a circular plate is 5.90 10-5 T · m2 when it is placed in a region of uniform magn

Question

The magnitude of the magnetic flux through the surface of a circular plate is 5.90 10-5 T · m2 when it is placed in a region of uniform magnetic field that is oriented at 42.0° to the vertical. The radius of the plate is 8.50 cm. Determine the strength of the magnetic field.

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Ngọc Khuê 3 months 2021-08-09T12:45:22+00:00 1 Answers 6 views 0

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    2021-08-09T12:46:35+00:00

    Answer:

    The strength of the magnetic field is 3.5 x 10⁻³ T

    Explanation:

    Given;

    magnitude of the magnetic flux , Φ = 5.90 x 10⁻⁵ T·m²

    angle of inclination of the field, θ = 42.0°

    radius of the circular plate, r = 8.50 cm = 0.085 m

    Generally magnetic flux in a uniform magnetic field is given as;

    Φ = BACosθ

    where;

    B is the strength of the magnetic field

    A is the area of the circular plate

    Area of the circular plate:

    A = πr²

    A = π (0.085)² = 0.0227 m²

    The strength of the magnetic field:

    B = Φ / ACosθ

    B = ( 5.90 x 10⁻⁵) / ( 0.0227 x Cos42)

    B = 3.5 x 10⁻³ T

    Therefore, the strength of the magnetic field is 3.5 x 10⁻³ T

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