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A 48.0-cm-diameter circular loop is rotated in a uniform electric field until the position of maximum electric flux is found. the flux in this position is measured to be 5.18 105 n · m2/c. what is the magnitude of the electric field?

Respuesta :

At the point of maximum flux, the electric field is directed normal to the plane of the circular loop. The electric flux may be calculated as, Φ = E * A Where we assume that the electric field, E, is uniform over the area of the circle, A. We may solve for an unknown electric field magnitude as, E = Φ / A Where we are told that the flux is: Φ = 5.18 E5 N m^2 / C, and we can calculate the area of the loop as, A = pi * (D/2)^2 = (3.1415) * (0.48 m / 2)^2 = .1809504 m^2 Therefore, E = (5.18 E5 N m^2 / C) / (.1809504 m^2) E = ~2.86 E6 N/C
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