P31_018

# P31_018 - πa 2 2 then the area(as it appears to us...

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18. (a) The rotational frequency (in revolutions per second) is identical to the time-dependent voltage frequency (in cycles per second, or Hertz). This conclusion should not be considered obvious, and the calculation shown in part (b) should serve to reinforce it. (b) First, we de±ne angle relative to the plane of Fig. 31-41, such that the semicircular wire is in the θ = 0 position and a quarter of a period (of revolution) later it will be in the θ = π/ 2 position (where its midpoint will reach a distance of a above the plane of the ±gure). At the moment it is in the θ = π/ 2 position, the area enclosed by the “circuit” will appear to us (as we look down at the ±gure) 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 θ =3 π/ 2 position). Since the area of the semicircle is
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Unformatted text preview: πa 2 / 2 then the area (as it appears to us) enclosed by the circuit, as a function of our angle θ , is A = A + πa 2 2 cos θ where (since θ is increasing at a steady rate) the angle depends linearly on time, which we can write either as θ = ωt or θ = 2 πft if we take t = 0 to be a moment when the arc is in the θ = 0 position. Since ~ B is uniform (in space) and constant (in time), Faraday’s law leads to E = − d Φ B dt = − B dA dt = − B d ³ A + πa 2 2 cos θ ´ dt = − B πa 2 2 d cos(2 πft ) dt which yields E = Bπ 2 a 2 f sin(2 πft ). This (due to the sinusoidal dependence) reinforces the conclu-sion in part (a) and also (due to the factors in front of the sine) provides the voltage amplitude: E max = Bπ 2 a 2 f ....
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## This note was uploaded on 11/12/2011 for the course PHYS 2001 taught by Professor Sprunger during the Fall '08 term at LSU.

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