QuestionDetails: Tags: Physics , Calculus BasedPhysics Tags: Physics , Calculus

Question

QuestionDetails: Tags: Physics, Calculus BasedPhysics Tags: Physics, Calculus BasedPhysics Tough Question Alert! Earn +8additional karma points for helping this annualmember. Tough Question Alert! Earn +8additional karma points for helping this annualmember. QuestionDetails: Consider the situation shown in the figure below. An electric fieldof 300 V/m is confined to a circular area d = 8.3 cm in diameter and directed outward perpendicularto the plane of the figure. If the field is increasing at a rate of17.8 V/m·s, whatare the direction and magnitude of the magnetic field at the pointP, 13.7 cm from the center of thecircle?

magnitude _______________T

Explanation / Answer

the magnetic force on the particle at point P is F = q * v * B or (m * v^2/r) = q * v * B or B = (m * v/q * r) ---------------(1) v = (2r/t) --------------(2) the rate of increase of electric field is (dE/dt) = 17.8 or dE = 17.8 * dt integrating both sides E = 17.8 * t or t = (E/17.8) -----------------(3) from (2) and (3) v = (2r/(E/17.8)) = (35.6r/E)-----------------(4) the electric potential is V = (1/4o) * (q/r) or q = V * 4o* r We know that V = E * d or q = E * d * 4o* r------------------(5) from (1),(4) and (5) B = (35.6 * m/4o* E^2 * d * r) m is mass of the electron and is equal to 9.1 * 10^-31kg o= 8.85 * 10^-12 C^2/Nm^2 d = 8.3 cm = 8.3 * 10^-2 m r = 13.7 cm = 13.7 * 10^-2 m the direction of the magnetic field is perpendicular to thedirection of electric field and therefore the magnetic field pointsout of the board.

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