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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?

Thanks you in advance.

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1 Answer 1

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You are correct. The induced emf will have the same polarity as that of the voltage source.

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  • $\begingroup$ Thanks. I will wait a bit for other answers before marking correct answer. $\endgroup$
    – deppep
    Jul 3, 2020 at 9:50

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