Induced Voltages and Inductance

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The discussion focuses on calculating the induced current in a resistor connected to horizontal rails as a rolling axle moves through a uniform magnetic field. Using Faraday's law, the induced electromotive force (E) is determined to be E = -B * L * u, leading to the induced current (I) calculated via Ohm's law as I = - (B * L * u) / R. The horizontal force required to maintain constant speed is also derived from the relationship between current, magnetic field, and force. The discussion also touches on the electric potential difference across the resistor and whether the current reverses direction after the axle passes the resistor. Overall, the calculations illustrate the principles of electromagnetism in motion.
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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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