Acceleration Ramp
About points...
We associate a certain number of points with each exercise.
When you click an exercise into a collection, this number will be taken as points for the exercise, kind of "by default".
But once the exercise is on the collection, you can edit the number of points for the exercise in the collection independently, without any effect on "points by default" as represented by the number here.
That being said... How many "default points" should you associate with an exercise upon creation?
As with difficulty, there is no straight forward and generally accepted way.
But as a guideline, we tend to give as many points by default as there are mathematical steps to do in the exercise.
Again, very vague... But the number should kind of represent the "work" required.
When you click an exercise into a collection, this number will be taken as points for the exercise, kind of "by default".
But once the exercise is on the collection, you can edit the number of points for the exercise in the collection independently, without any effect on "points by default" as represented by the number here.
That being said... How many "default points" should you associate with an exercise upon creation?
As with difficulty, there is no straight forward and generally accepted way.
But as a guideline, we tend to give as many points by default as there are mathematical steps to do in the exercise.
Again, very vague... But the number should kind of represent the "work" required.
About difficulty...
We associate a certain difficulty with each exercise.
When you click an exercise into a collection, this number will be taken as difficulty for the exercise, kind of "by default".
But once the exercise is on the collection, you can edit its difficulty in the collection independently, without any effect on the "difficulty by default" here.
Why we use chess pieces? Well... we like chess, we like playing around with \(\LaTeX\)-fonts, we wanted symbols that need less space than six stars in a table-column... But in your layouts, you are of course free to indicate the difficulty of the exercise the way you want.
That being said... How "difficult" is an exercise? It depends on many factors, like what was being taught etc.
In physics exercises, we try to follow this pattern:
Level 1 - One formula (one you would find in a reference book) is enough to solve the exercise. Example exercise
Level 2 - Two formulas are needed, it's possible to compute an "in-between" solution, i.e. no algebraic equation needed. Example exercise
Level 3 - "Chain-computations" like on level 2, but 3+ calculations. Still, no equations, i.e. you are not forced to solve it in an algebraic manner. Example exercise
Level 4 - Exercise needs to be solved by algebraic equations, not possible to calculate numerical "in-between" results. Example exercise
Level 5 -
Level 6 -
When you click an exercise into a collection, this number will be taken as difficulty for the exercise, kind of "by default".
But once the exercise is on the collection, you can edit its difficulty in the collection independently, without any effect on the "difficulty by default" here.
Why we use chess pieces? Well... we like chess, we like playing around with \(\LaTeX\)-fonts, we wanted symbols that need less space than six stars in a table-column... But in your layouts, you are of course free to indicate the difficulty of the exercise the way you want.
That being said... How "difficult" is an exercise? It depends on many factors, like what was being taught etc.
In physics exercises, we try to follow this pattern:
Level 1 - One formula (one you would find in a reference book) is enough to solve the exercise. Example exercise
Level 2 - Two formulas are needed, it's possible to compute an "in-between" solution, i.e. no algebraic equation needed. Example exercise
Level 3 - "Chain-computations" like on level 2, but 3+ calculations. Still, no equations, i.e. you are not forced to solve it in an algebraic manner. Example exercise
Level 4 - Exercise needs to be solved by algebraic equations, not possible to calculate numerical "in-between" results. Example exercise
Level 5 -
Level 6 -
Question
Solution
Short
Video
\(\LaTeX\)
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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 an 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
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 an 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
Meta Information
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 an 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
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 an 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
Contained in these collections:
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Magnetic Force (BC) by by
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Magnetic Forces by by

