# a metal bar on a pair of conducting rails that carries a current

This is a homework question, and I solved it already, but something bugs me. So the problem is stated as following:

A metal bar of mass M sits on a pair of long horizontal conducting rails separated by a distance $\mathcal{l}$ and connected to a device that supplies a constant I to the circuit, also the bar has length l. There is a constant magnetic field $\vec{B}$ that points into the screen where this whole setup is. The current is following in clockwise direction.

Now if there is no friction on the rail, the bar will move to the right because of $$F = I\vec{L} \times \vec{B}$$ This is easy.

And since there is no friction, the acceleration will be $$a = \dfrac{F}{M}$$ and hence the magnitude of the velocity at any later time $t$ will be $$v(t) = at = \dfrac{IlB}{M}t$$

The question is asking for the velocity at time $t$.

However, what bugs me is that since the bar is moving to the right, thus the flux through the closed loop of the powersource, rails and the bar will be increasing, which means that there will be an induced electric potential trying to push back the current I. However, the problem does not give the resistance of the bar, and I assume that in this hypothetical situation, the whole thing has 0 resistance. So how does the electric potential make sense in this case? Since I am not given a resistance, I cannot calculate the "pushing back" current caused by the change in flux. Thank you!

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You're right that there is an induced EMF pushing against the current. In order to keep this going, sufficient voltage must be applied to overcome this induced EMF, thereby maintaining a constant current. Since you are given a constant current source, you can assume (as if by magic) that this condition is satisfied. As long as the applied voltage is sufficient to overcome the back-EMF and maintain a constant current, the bar continues to accelerate. This may lead to complicated requirements on the applied voltage, however.

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I see. Thank you for your explanation! – Enzo Mar 1 '13 at 15:30