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antbanks3050
26.03.2020 •
Physics
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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Ответ:
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
Ответ: