Induced Voltages and Inductance

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SUMMARY

The discussion focuses on the calculation of induced current and forces in a closed-loop circuit formed by a rolling axle and rails in a uniform magnetic field. Given the parameters, the induced current I can be calculated using the formula i = -(B * L * u) / R, where B = 0.0800 T, L = 1.50 m, u = 3.00 m/s, and R = 0.400 Ω. The horizontal force F required to maintain constant speed can be derived from the induced current and the magnetic field. Additionally, the potential difference across the resistor indicates that point a is at a higher electric potential than point b. The current does not reverse direction after the axle rolls past the resistor.

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In the figure attached below, the rolling axle, 1.50 m long, is pushed along horizontal rails at a constant speed u = 3.00 m/s. A resistor R= 0.400 Ω is connected to the rails at points a and b, directly opposite each other. (The wheels make good electrical contact with the rails, and so the axle, rails, and R form a closed-loop circuit. The only significant resistance in the circuit is R.) There is a uniform magnetic field B= 0.0800 T vertically downward.

a)Find the induced current I in the resistor.
b)What horizontal force F is required to keep the axle rolling at constant speed?
c)Which end of the resistor, a or b, is at the higher electric potential?
d)After the axle rolls past the resistor, does the current in R reverse direction?
 

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First use Faraday's law E=-N d(BA)/dx to find the voltage, and just take into consideration that B=constant. Therefore, E=-N Bd(A)/dx and also A=l(variable). Then use ohms law (E=iR) and these three other equations (F=iL x B, P=Fu and P=i^2R). I'm sorry I've almost forgot N=1.
 
I apologize to everybody for writing E=-Nd(BA)/dx. Instead, I've should written
E=-Nd(BA)/dt and because of that I;m going to solve a).
E=-1(B)d(A)/dt=-B(l)dx/dt=-B(l)(u)=-B*L*u and from ohms law i=-(B*L*u)/R
 

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