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17 May 2013

EMF induced in a straight wire -- 2013 AP Physics B problem 6(e)

I feel a disturbance in the force... problem 6 on the 2013 AP Physics B exam seems to be generating significant discussion of emf induced in a straight wire, rather than in a loop of wire.

First of all, take a look at the question, to be found in this file at  I can't post the question here for lawyerly reasons.    

In the last part, we have a wire carrying a current to the right.  A second wire is placed above the first wire, and is moved upward at constant speed.  

Finding the amount of induced emf is easy: ε = BLv.  Although the magnetic field is different at different distances from the current-carrying wire, the question asks about induced emf at one specific location.  So calculate the magnetic field B produced there, use the length of the wire, and the speed the wire is moving.  Fine.

But which side of the wire is at a higher electric potential?  Lenz's law is the usual approach to such questions.  Without a loop of wire, though, I don't know how to apply my usual approach to Lenz's law.

The way I understand this situation is to consider the effect of the magnetic field on positive charge carriers in the wire.*  The force on these positive charge carriers is given by the right hand rule associated with F = qvB, the magnetic force on a moving charge.  Which direction is this magnetic force?

*Or on the electrons, or both, or whatever.  When you've got a conductor, you can equivalently discuss the motion of electrons or the "holes in the electron sea" which are essentially "positive charge carriers."  Who cares.  Just don't call 'em "protons" and we're all copacetic.

The positive charge carriers are moving with the wire, or up the page.  The magnetic field produced by the rightward current in the bottom wire is out of the page (by a different right hand rule).  Thus, the force on these positive charge carriers is to the right.  Similarly, the force on negative charge carriers in the moving wire is to the left.

So as the wire moves, the right end of the wire becomes more positively charged, the left end becomes more negatively charged.  Which is at the higher electric potential?

The right side is at a higher electric potential.

"Whoa there, Hoss," you say.  "By your own definition, positive charges are forced from high to low electric potential.  These positive charges were forced to the right... so why is the right side the HIGHER electric potential?"

Because the positive charge carriers were forced to the right by a magnetic force, not an electric force.

To find the side with higher electric potential, consider what would happen ELECTRICALLY to a positive test charge placed in the wire, ignoring magnetic effects.  This positive test charge would be repelled electrostatically by the positive right side, and attracted to the negative left side.  Positive test charges are forced from high to low electric potential; the right side must be at higher electric potential.

A more conceptual approach:  Again, separate magnetic and electric effects.  The right side becomes positively charge, the left side negative.  Now if the magnetic effects were removed, which way would the current -- flow of positive charge -- be?  Obviously from + to -, or from right to left.  Current flows out of the positive side of a battery, or from the high voltage to low voltage.  So the right end is at higher "voltage."


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