# Conducting rod on sliding rails. Is this book's solution right?

Dears all,

I have the following exercise:

A sliding rod of length $$b = 0.2 m$$ is placed above two conducting rails connected to a voltage generator $$V_0 = 6V$$ (see fig). The rod has a resistance $$R = 0.08 \Omega$$ and is attached to a mass $$m = 1.2 kg$$ through a pulley. The entire system is exposed to a magnetic field perpendicular the circuit plane and pointing out of the picture, with intensity $$B=1T$$. Calculate the limit velocity of the bar and the current flowing through the circuit in such circumstances.

The exercise is solved imposing the following equation:

1. the acceleration of the mass attached to the sliding rod is determined by weight and Lorentz's force acting on the bar via $$ma = mg - IBv$$

2. the current flowing into the circuit is determined by the generator voltage and the Faraday law.

Now I have a problem with the second equation. Assuming the mass is enough to pull the rod to the left, the enclosed flux decreases hence causing a current with same polarity of the generator (flowing counterclockwise): $$RI = V_0 + bBv$$.

However the exercise's solution (which come from an old book) reports: $$RI = V_0 - bBv$$.

Is the book's solution wrong or do I have a misunderstanding with Faraday's Law?