ch30-p015 - 15. (a) The frequency is f= (40 rev/s)(2...

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θ = π /2 position (where its midpoint will reach a distance of a above the plane of the figure). At the moment it is in the = π /2 position, the area enclosed by the “circuit” will appear to us (as we look down at the figure) to that of a simple rectangle (call this area A 0 which is the area it will again appear to enclose when the wire is in the = 3 π /2 position). Since the area of the semicircle is π a 2 /2 then the area (as it appears to us) enclosed by the circuit, as a function of our angle , is AA a =+ 0 2 2 π cos where (since is increasing at a steady rate) the angle depends linearly on time, which we can write either as = ω t or = 2 π ft if we take t = 0 to be a moment when the arc is in the = 0 position. Since G B is uniform (in space) and constant (in time), Faraday’s law
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This note was uploaded on 06/03/2011 for the course PHY 2049 taught by Professor Any during the Spring '08 term at University of Florida.

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