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 ....
View Full Document
This note was uploaded on 11/12/2011 for the course PHYS 2001 taught by Professor Sprunger during the Fall '08 term at LSU.
- Fall '08