The force on the electronsinside the metal bar is F = -q e v ´ B . Here v ´ B po

Question

The force on the electronsinside the metal bar is F = -qev ´ B. Herev ´ B points into the page,so the force on the electrons is out of the page and has magnitudeqevB. Electrons will move towards theoutward-facing surface of the bar and leave excess positive chargeon the inward-facing surface. This separated charge producesan electric field E, which exerts a forceqeE on the electrons in the oppositedirection.   E points out of the pageand exerts a force directed into the page on the electrons. When qeE = qevB the net force is zero. Thenequilibrium has been reached. The voltage between theoutward-facing and inward-facing surface is Ed, where d thethickness of the bar. The inward-facing (positive) side is atthe higher potential.

There is no potentialdifference between ends 1 and 2.

Explanation / Answer

A)a potential difference between its ends, end1 being at a higher potential. B)a potential difference between its ends, end2 being at a higher potential. C)no potentialdifference between its ends.    (because themagnetic flux through the surface remains constant) D)a time varying potential difference acrossits ends. E)none of the above.

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