Energiedichte von elektromagnetischen Feldern
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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Exercise:
Zeige aus den Maxwell-Gleichungen im Vakuum dass elektromagnetische Felder einem Energieerhaltungssatz gehorchen: fracpartial wpartial t + nabla vecS quadtextmitquad w frac varepsilon_ E^ + frac fracB^mu_ quad vecS fracmu_ vecE times vecB Diese Gleichung stellt den Energieerhaltungssatz für elektromagnetische Felder dar.
Solution:
Wir nen mit den Maxwell-Gleichungen im Vakuum: * nabla vecE nabla vecB nabla times vecE -fracpartial vecBpartial t nabla times vecB mu_ varepsilon_ fracpartial vecEpartial t * Wir betrachten das Skalarprodukt von vecE mit der vierten Gleichung: * vecE nabla times vecB mu_ varepsilon_ vecE fracpartial vecEpartial t * und das Skalarprodukt von vecB mit der dritten Gleichung: * vecB nabla times vecE - vecB fracpartial vecBpartial t * Nun subtrahieren wir die beiden Gleichungen: * vecE nabla times vecB - vecB nabla times vecE mu_ varepsilon_ vecE fracpartial vecEpartial t + vecB fracpartial vecBpartial t * Auf der linken Seite steht nach einer Vektoridentität: nabla vecE times vecB vecB nabla times vecE - vecE nabla times vecB - left vecE nabla times vecB - vecB nabla times vecE right Somit ergibt sich: nabla vecE times vecB - mu_ varepsilon_ vecE fracpartial vecEpartial t - vecB fracpartial vecBpartial t Nun erkennen wir: vecE fracpartial vecEpartial t frac fracpartialpartial t E^ quad vecB fracpartial vecBpartial t frac fracpartialpartial t B^ Einsetzen ergibt: nabla vecE times vecB - frac mu_ varepsilon_ fracpartial E^partial t - frac fracpartial B^partial t Multiplizieren mit fracmu_ : fracmu_ nabla vecE times vecB - frac varepsilon_ fracpartial E^partial t - fracmu_ fracpartial B^partial t Nun definieren wir: w frac varepsilon_ E^ + fracmu_ B^ quad vecS fracmu_ vecE times vecB Daraus ergibt sich der Energieerhaltungssatz: boxed fracpartial wpartial t + nabla vecS
Zeige aus den Maxwell-Gleichungen im Vakuum dass elektromagnetische Felder einem Energieerhaltungssatz gehorchen: fracpartial wpartial t + nabla vecS quadtextmitquad w frac varepsilon_ E^ + frac fracB^mu_ quad vecS fracmu_ vecE times vecB Diese Gleichung stellt den Energieerhaltungssatz für elektromagnetische Felder dar.
Solution:
Wir nen mit den Maxwell-Gleichungen im Vakuum: * nabla vecE nabla vecB nabla times vecE -fracpartial vecBpartial t nabla times vecB mu_ varepsilon_ fracpartial vecEpartial t * Wir betrachten das Skalarprodukt von vecE mit der vierten Gleichung: * vecE nabla times vecB mu_ varepsilon_ vecE fracpartial vecEpartial t * und das Skalarprodukt von vecB mit der dritten Gleichung: * vecB nabla times vecE - vecB fracpartial vecBpartial t * Nun subtrahieren wir die beiden Gleichungen: * vecE nabla times vecB - vecB nabla times vecE mu_ varepsilon_ vecE fracpartial vecEpartial t + vecB fracpartial vecBpartial t * Auf der linken Seite steht nach einer Vektoridentität: nabla vecE times vecB vecB nabla times vecE - vecE nabla times vecB - left vecE nabla times vecB - vecB nabla times vecE right Somit ergibt sich: nabla vecE times vecB - mu_ varepsilon_ vecE fracpartial vecEpartial t - vecB fracpartial vecBpartial t Nun erkennen wir: vecE fracpartial vecEpartial t frac fracpartialpartial t E^ quad vecB fracpartial vecBpartial t frac fracpartialpartial t B^ Einsetzen ergibt: nabla vecE times vecB - frac mu_ varepsilon_ fracpartial E^partial t - frac fracpartial B^partial t Multiplizieren mit fracmu_ : fracmu_ nabla vecE times vecB - frac varepsilon_ fracpartial E^partial t - fracmu_ fracpartial B^partial t Nun definieren wir: w frac varepsilon_ E^ + fracmu_ B^ quad vecS fracmu_ vecE times vecB Daraus ergibt sich der Energieerhaltungssatz: boxed fracpartial wpartial t + nabla vecS
Meta Information
Exercise:
Zeige aus den Maxwell-Gleichungen im Vakuum dass elektromagnetische Felder einem Energieerhaltungssatz gehorchen: fracpartial wpartial t + nabla vecS quadtextmitquad w frac varepsilon_ E^ + frac fracB^mu_ quad vecS fracmu_ vecE times vecB Diese Gleichung stellt den Energieerhaltungssatz für elektromagnetische Felder dar.
Solution:
Wir nen mit den Maxwell-Gleichungen im Vakuum: * nabla vecE nabla vecB nabla times vecE -fracpartial vecBpartial t nabla times vecB mu_ varepsilon_ fracpartial vecEpartial t * Wir betrachten das Skalarprodukt von vecE mit der vierten Gleichung: * vecE nabla times vecB mu_ varepsilon_ vecE fracpartial vecEpartial t * und das Skalarprodukt von vecB mit der dritten Gleichung: * vecB nabla times vecE - vecB fracpartial vecBpartial t * Nun subtrahieren wir die beiden Gleichungen: * vecE nabla times vecB - vecB nabla times vecE mu_ varepsilon_ vecE fracpartial vecEpartial t + vecB fracpartial vecBpartial t * Auf der linken Seite steht nach einer Vektoridentität: nabla vecE times vecB vecB nabla times vecE - vecE nabla times vecB - left vecE nabla times vecB - vecB nabla times vecE right Somit ergibt sich: nabla vecE times vecB - mu_ varepsilon_ vecE fracpartial vecEpartial t - vecB fracpartial vecBpartial t Nun erkennen wir: vecE fracpartial vecEpartial t frac fracpartialpartial t E^ quad vecB fracpartial vecBpartial t frac fracpartialpartial t B^ Einsetzen ergibt: nabla vecE times vecB - frac mu_ varepsilon_ fracpartial E^partial t - frac fracpartial B^partial t Multiplizieren mit fracmu_ : fracmu_ nabla vecE times vecB - frac varepsilon_ fracpartial E^partial t - fracmu_ fracpartial B^partial t Nun definieren wir: w frac varepsilon_ E^ + fracmu_ B^ quad vecS fracmu_ vecE times vecB Daraus ergibt sich der Energieerhaltungssatz: boxed fracpartial wpartial t + nabla vecS
Zeige aus den Maxwell-Gleichungen im Vakuum dass elektromagnetische Felder einem Energieerhaltungssatz gehorchen: fracpartial wpartial t + nabla vecS quadtextmitquad w frac varepsilon_ E^ + frac fracB^mu_ quad vecS fracmu_ vecE times vecB Diese Gleichung stellt den Energieerhaltungssatz für elektromagnetische Felder dar.
Solution:
Wir nen mit den Maxwell-Gleichungen im Vakuum: * nabla vecE nabla vecB nabla times vecE -fracpartial vecBpartial t nabla times vecB mu_ varepsilon_ fracpartial vecEpartial t * Wir betrachten das Skalarprodukt von vecE mit der vierten Gleichung: * vecE nabla times vecB mu_ varepsilon_ vecE fracpartial vecEpartial t * und das Skalarprodukt von vecB mit der dritten Gleichung: * vecB nabla times vecE - vecB fracpartial vecBpartial t * Nun subtrahieren wir die beiden Gleichungen: * vecE nabla times vecB - vecB nabla times vecE mu_ varepsilon_ vecE fracpartial vecEpartial t + vecB fracpartial vecBpartial t * Auf der linken Seite steht nach einer Vektoridentität: nabla vecE times vecB vecB nabla times vecE - vecE nabla times vecB - left vecE nabla times vecB - vecB nabla times vecE right Somit ergibt sich: nabla vecE times vecB - mu_ varepsilon_ vecE fracpartial vecEpartial t - vecB fracpartial vecBpartial t Nun erkennen wir: vecE fracpartial vecEpartial t frac fracpartialpartial t E^ quad vecB fracpartial vecBpartial t frac fracpartialpartial t B^ Einsetzen ergibt: nabla vecE times vecB - frac mu_ varepsilon_ fracpartial E^partial t - frac fracpartial B^partial t Multiplizieren mit fracmu_ : fracmu_ nabla vecE times vecB - frac varepsilon_ fracpartial E^partial t - fracmu_ fracpartial B^partial t Nun definieren wir: w frac varepsilon_ E^ + fracmu_ B^ quad vecS fracmu_ vecE times vecB Daraus ergibt sich der Energieerhaltungssatz: boxed fracpartial wpartial t + nabla vecS
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