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This comes from one of the free response questions from the AP Physics B 2014 exam.

diagram of rod suspended by springs in magnetic field

5. (15 points) A conducting rod of mass $m$ and length $L$ hangs at rest from two identical conducting springs, each with spring constant $k$, as shown in the figure at left above. The upper ends of the springs are fixed at points $P$ and $Q$, and the rod is in a uniform magnetic field $\mathbf{B}$ directed into the page. A battery is then connected between points $P$ and $Q$, as shown in the figure at right above, resulting in a current $I$ in the rod. The rod is displaced downward, eventually reaching a new equilibrium position with the springs stretched an additional distance $\Delta y$.

(a) Which point, $P$ or $Q$, is connected to the positive terminal of the battery? Justify your answer.

The answer is Q. But I put P. I said there will be a change and increase in flux. This means, by Lenz's Law, that there will be an induced magnetic field in the opposite direction. Using the hand rule, you point your thumb in the direction of the magnetic field, and CURL your fingers towards the direction of current, which makes it counter-clockwise. Thus, it can be seen that P is the right answer.

Why is my answer wrong? My other classmates are telling me the answer is Q, why?

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3 Answers 3

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The answer is Q because you can tell from the second picture that a downward force is being applied to the rod when the battery is connected. Because the field is directed into the page the current has to be moving from point Q to point P to create this downward force.

It's true that in moving from the initial to final position, there will be an induced current because of the change in flux, but this is a transient effect which no longer occurs once the rod reaches equilibrium, and I assume that the current created by the battery dwarfs this transient effect anyway.

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  • $\begingroup$ okay, so the current created by the battery opposes the induced current created by the change in flux? Overall, this results in no current flowing into the loop when it reaches equilibrium? $\endgroup$
    – Varun Iyer
    May 14, 2014 at 23:00
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    $\begingroup$ No, once the loop reaches equilibrium the current is not zero. Once the loop reaches equilibrium there is no longer a changing flux, meaning there is no more induced current. That's what I meant when I said the induced current is a transient effect. Once the loop is no longer moving the current is supplied solely by the battery. $\endgroup$
    – wgrenard
    May 15, 2014 at 2:21
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Lenz's law is the wrong physical principle to use here. It applies when you have a change in magnetic flux that causes a current, but in this case, it's the other way around: the current causes the change in flux by making the bar move. Because the current comes from the battery, not from a changing magnetic flux, you don't use Lenz's law on it.

Now, as the bar moves, there is a change in flux, and that change in flux will cause an induced current on top of the current from the battery. Actually, the induced current flows in the opposite direction from the original current, so it cancels out a little bit of the original current. (The induced current is generally much less than the original current.) But the induced current doesn't matter for this problem; you only need to know about the original current.

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Since the rod has been pushed downwards due to magnetic interaction, we can figure out using $ F = I L×B $ that the current goes from Q to P.

Now if you want to apply lenz laws you would find that an increasing flux would induce a current that goes from P to Q, but that is an induced current and what is asked of us is to tell which is the positive terminal of the battery. This has been determined in previous paragraph to be Q only!

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