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Exercise:
The two conducting rails in the drawing are tilted upward so they each make an angle of alO with respect to the ground. The vertical magnetic field has a magnitude of BO. The mO aluminum rod length LO slides without friction down the rails at a constant velocity. includegraphicswidth.textwidth#image_path:accelerated-rod# abcliste abc Determine the direction of the current. abc How much current flows through the rod? abc Calculate the acceleration for a current of IbO. abcliste

Solution:
abcliste abc The two relevant forces are the components F_Gparallel of the gravitational force and F_Bparallel of the magnetic force parallel to the rails. F_Bparallel has to oppose the effect of F_Gparallel i.e. the magnetic force has to have a component in the upward direction. This is only possible if the current flows from right to left. abc The horizontal magnetic force is given by F_B I L B The component F_Bparallel parallel to the rails is F_Bparallel F_B cosalpha I L B cosalpha The component F_Gparallel is F_Gparallel F_Gsinalpha m gsinalpha For a constant velocity the two components have to cancel each other. It follows for the current I fracm gsinalphaL Bcosalhpa IF fracmtimes ncgLtimes Btimestanal I approx resultIP abc The resultant force parallel to the rails is given by sscFres F_Bparallel-F_Gparallel I L B cosalpha - m gsinalpha Using Newton's second law we find for the acceleration a fracsscFresm aF fracIbtimes Ltimes Btimes cosalm - ncgtimessinal a approx resultaP abcliste
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Exercise:
The two conducting rails in the drawing are tilted upward so they each make an angle of alO with respect to the ground. The vertical magnetic field has a magnitude of BO. The mO aluminum rod length LO slides without friction down the rails at a constant velocity. includegraphicswidth.textwidth#image_path:accelerated-rod# abcliste abc Determine the direction of the current. abc How much current flows through the rod? abc Calculate the acceleration for a current of IbO. abcliste

Solution:
abcliste abc The two relevant forces are the components F_Gparallel of the gravitational force and F_Bparallel of the magnetic force parallel to the rails. F_Bparallel has to oppose the effect of F_Gparallel i.e. the magnetic force has to have a component in the upward direction. This is only possible if the current flows from right to left. abc The horizontal magnetic force is given by F_B I L B The component F_Bparallel parallel to the rails is F_Bparallel F_B cosalpha I L B cosalpha The component F_Gparallel is F_Gparallel F_Gsinalpha m gsinalpha For a constant velocity the two components have to cancel each other. It follows for the current I fracm gsinalphaL Bcosalhpa IF fracmtimes ncgLtimes Btimestanal I approx resultIP abc The resultant force parallel to the rails is given by sscFres F_Bparallel-F_Gparallel I L B cosalpha - m gsinalpha Using Newton's second law we find for the acceleration a fracsscFresm aF fracIbtimes Ltimes Btimes cosalm - ncgtimessinal a approx resultaP abcliste
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Attributes & Decorations
Branches
Kinematics, Magnetism
Tags
acceleration, electromagnetism, force, lorentzkraft, magnetism, mechanics, mechanik, physics
Content image
Difficulty
(3, default)
Points
3 (default)
Language
ENG (English)
Type
Calculative / Quantity
Creator by
Decoration