Parallel Currents
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
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\(\LaTeX\)
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Exercise:
Two straight parallel wires are placed at a distance of dO. They carry the currents IaO and IbO respectively opposite directions. Calculate the magnitude of the magnetic field at a po in the plane defined by the wires which abcliste abc has the same distance to both wires; abc has distance cm from the first and cm from the second wire. abc Find the pos in the plane where the magnetic field vanishes. abcliste
Solution:
abcliste abc The partial fields produced by the currents po in the same direction so the contributions can be added: B B_+B_ fracmu_pi leftfracI_r_+fracI_r_right labelrefB For a po in the middle between the two wires r_ r_ d/ we find B BaF fracncmuopitimes dtimes Ia+Ib Ba approx resultBaP- abc Using refrefB we find B fracncmuopitimes leftfracIara+fracIbrbright Bb approx resultBbP- abc In order for the two partial fields to cancel at some po the po is not allowed to be between the wires. It is also clear that it has to be closer to the weaker current. This means that r_r_+d. The magnitudes of B_ and B_ must be equal: B_ fracmu_pifracI_r_ fracmu_pifracI_r_ B_ This can be simplified to fracI_r_ fracI_r_ fracI_r_+d It follows that I_ r_+d I_ r_+I_ d I_ r_ Solving for r_ leads to r_ rF dOtimes fracIaIb-Ia resultrP- The combined magnetic field disappears on a line parallel to the two wires in the plane defined by the wires. It is rP- away from the weaker current in the direction away from the stronger current. abcliste
Two straight parallel wires are placed at a distance of dO. They carry the currents IaO and IbO respectively opposite directions. Calculate the magnitude of the magnetic field at a po in the plane defined by the wires which abcliste abc has the same distance to both wires; abc has distance cm from the first and cm from the second wire. abc Find the pos in the plane where the magnetic field vanishes. abcliste
Solution:
abcliste abc The partial fields produced by the currents po in the same direction so the contributions can be added: B B_+B_ fracmu_pi leftfracI_r_+fracI_r_right labelrefB For a po in the middle between the two wires r_ r_ d/ we find B BaF fracncmuopitimes dtimes Ia+Ib Ba approx resultBaP- abc Using refrefB we find B fracncmuopitimes leftfracIara+fracIbrbright Bb approx resultBbP- abc In order for the two partial fields to cancel at some po the po is not allowed to be between the wires. It is also clear that it has to be closer to the weaker current. This means that r_r_+d. The magnitudes of B_ and B_ must be equal: B_ fracmu_pifracI_r_ fracmu_pifracI_r_ B_ This can be simplified to fracI_r_ fracI_r_ fracI_r_+d It follows that I_ r_+d I_ r_+I_ d I_ r_ Solving for r_ leads to r_ rF dOtimes fracIaIb-Ia resultrP- The combined magnetic field disappears on a line parallel to the two wires in the plane defined by the wires. It is rP- away from the weaker current in the direction away from the stronger current. abcliste
Meta Information
Exercise:
Two straight parallel wires are placed at a distance of dO. They carry the currents IaO and IbO respectively opposite directions. Calculate the magnitude of the magnetic field at a po in the plane defined by the wires which abcliste abc has the same distance to both wires; abc has distance cm from the first and cm from the second wire. abc Find the pos in the plane where the magnetic field vanishes. abcliste
Solution:
abcliste abc The partial fields produced by the currents po in the same direction so the contributions can be added: B B_+B_ fracmu_pi leftfracI_r_+fracI_r_right labelrefB For a po in the middle between the two wires r_ r_ d/ we find B BaF fracncmuopitimes dtimes Ia+Ib Ba approx resultBaP- abc Using refrefB we find B fracncmuopitimes leftfracIara+fracIbrbright Bb approx resultBbP- abc In order for the two partial fields to cancel at some po the po is not allowed to be between the wires. It is also clear that it has to be closer to the weaker current. This means that r_r_+d. The magnitudes of B_ and B_ must be equal: B_ fracmu_pifracI_r_ fracmu_pifracI_r_ B_ This can be simplified to fracI_r_ fracI_r_ fracI_r_+d It follows that I_ r_+d I_ r_+I_ d I_ r_ Solving for r_ leads to r_ rF dOtimes fracIaIb-Ia resultrP- The combined magnetic field disappears on a line parallel to the two wires in the plane defined by the wires. It is rP- away from the weaker current in the direction away from the stronger current. abcliste
Two straight parallel wires are placed at a distance of dO. They carry the currents IaO and IbO respectively opposite directions. Calculate the magnitude of the magnetic field at a po in the plane defined by the wires which abcliste abc has the same distance to both wires; abc has distance cm from the first and cm from the second wire. abc Find the pos in the plane where the magnetic field vanishes. abcliste
Solution:
abcliste abc The partial fields produced by the currents po in the same direction so the contributions can be added: B B_+B_ fracmu_pi leftfracI_r_+fracI_r_right labelrefB For a po in the middle between the two wires r_ r_ d/ we find B BaF fracncmuopitimes dtimes Ia+Ib Ba approx resultBaP- abc Using refrefB we find B fracncmuopitimes leftfracIara+fracIbrbright Bb approx resultBbP- abc In order for the two partial fields to cancel at some po the po is not allowed to be between the wires. It is also clear that it has to be closer to the weaker current. This means that r_r_+d. The magnitudes of B_ and B_ must be equal: B_ fracmu_pifracI_r_ fracmu_pifracI_r_ B_ This can be simplified to fracI_r_ fracI_r_ fracI_r_+d It follows that I_ r_+d I_ r_+I_ d I_ r_ Solving for r_ leads to r_ rF dOtimes fracIaIb-Ia resultrP- The combined magnetic field disappears on a line parallel to the two wires in the plane defined by the wires. It is rP- away from the weaker current in the direction away from the stronger current. abcliste
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