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Cont. Error in today's science

 Now, I am going to apply work using Prescotts on an electrical circuit. ***************************Let's find the average drift velocity: ------------------------- A is the average (weighted with respect to L) cross-section of the wire (m²) n is "free" electrons per unit volume (electrons/m³) e is the magnitude of charge of an electron (1.602 * 10^19 C/electron) v is the average drift velocity of the electrons (m/s) I is the current in the (C/s) dq is an infinitesmal amount of charge (C) dt is an infinitesmal amount of time (s) dN is an infinitesmal number of electrons (electrons) ------------------------- (1) dq = e*dN dN = nAv*dt (2) dt = dN/(nAv) (1)/(2) dq/dt = e*dN/(dN/nAv) I = enAv v = I/(enA)***************************Let's find force: ------------------------- W_j is the Work in Joules (N*m) f is the force (N) s is the distance (m) V is volts (N*m/C) ------------------------- W_j = F*s dW_j = F*v*dt dW_j/dt = F*v V*I = F*v V*I F = ----- v = VenA ------------------------- P is pressure (Pa) -------------------------So, F V = --- enA P = -- en We can now omit the use of Joules in the description of Volts. We can say that "Voltage is the electromagnetic-pressure (created by an EMF source) per density of charge."Notice that the pressure supplied by an EMF has nothing to do with the length of the circuit. A battery hooked to a 1 meter circuit of 1cm² wire uses the same force as a similar battery hooked to a 100 meter circuit of similar wire! Yet, it's obvious that more *work* is being done in the 100 meter circuit than in the 1 meter circuit. The reason why the force is the same while the work isn't is not hard at all to understand. An EMF source creates "electromagnetic pressure" on the anode and/or cathode. Once a circuit is started, this electromagnetic pressure is felt throughout the circuit. You can imagine the electrons as being dominoes. Whether you have 1 meter of "dominoes" falling or 100 meters of "dominoes" falling, the initial pressure or force may be the same, and yet, the amount of work done can be very different. (This obviously means that energy *isn't* conserved. That's right.)*************************** ------------------------- W_p is the Work in Prescotts (N*s) t is a duration of time (s) ------------------------- W_p = F*t = VenA*t *************************** ------------------------- U is Work (in Prescotts) per Coulomb (N*s/C) Q is an amount of charge (C) p is the resistivity of the wire (ohms) L is the length of the wire (m) ------------------------- U = W_p/Q = F/I = (VenA)/(V/R) = enAR = enA*(p*L/A) = enpLNow, U is a constant for any given circuit. So, given any circuit, it takes a constant amount of work to move a Coulomb along the circuit. Makes sense that it doesn't vary..*************************** ------------------------- µ is Work (in Prescotts) per Coulomb meter (N*s/(C*m)) ------------------------- µ = dU/dL = enpThus, the rate at which work is done per unit distance depends only on the material. Makes sense..*************************** ------------------------- t_c is the average change in time between electron collisions (s) m_e is the mass of an electron (9.109 * 10^(-31) kg/electron) -------------------------Each electron gains m_e*2v of energy before it makes a collision and losses it's energy. The collision will take place in t_c seconds. U is the amount of work to move a Coulomb L meters. Thus, in L meters, there will be L/(v*t_c) number of collisions. So, L m_e*2v ----- * ------ = enpL v*t_c e 2m_e t_c = ---- e²np which is correct.---------------------------------------------   Message Thread Cont. Error in today's science - Raheman Velji 11/20/2003, 7:03 pm « Back to index | View thread » 