# Solution consider each statement above 1 false at

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Solution:Consider each statement above:(1) False At position P themagneticforce is upward and the current in the ring is clock-wise.
(1) Commentary The direction of the motional emf can be deduced from the expressionE=H~v×~B.In this expression only the velocity of the part of the ring that isactually within the magnetic field matters (sinceB= 0 for the parts outside ofthe field). The velocity of the ring points downward throughout its entire motion(because it is falling) so~v×~Bwillalwayspoint to the right. At position P theemf points to the right in thebottompart of the ring, so the emf (and thecurrent) will be counter-clockwise. The magnetic force on the ring depends uponthe direction of the current for the part of the ring inside of the magnetic field.As we just determined, the direction of the current is to the right in the bottomof the ring, so~F=I(d~l)×~Bwill point up.(2) False At position S thenetforce (~Fnet=~FB+~Fg) is zero and the current in the ringis counter-clockwise.(2) Commentary The entire ring is inside of the magnetic field. The emf along the bottom of thering and the emf along the top of the ring both point to the right (because~v×~Bpoint right everywhere) so the total emf around the ring is zero.This meansthat there is no current and no magnetic force on the ring. You can also arriveat this conclusion from the fact that the magnetic flux through the loop is notchanging. (There is still a gravitational force, so the net force still points down.)(3) True At position R themagneticforce is upward and the current in the ring is clock-wise.(3) Commentary Only the top of the ring is within the magnetic field here. This means that theemf points to the right in the top part, and thus creates a clockwise current. Thecurrent in the top part of the ring points to the right, and the magnetic forcewill be upward.

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Term
Fall
Professor
Evrard
Tags
Magnetic Field, LRC, prms, Prof David Gerdes