Magnetic field in Zurich
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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
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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 -
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Exercise:
For the magnetic field in Zurich look up the horizontal component the inclination and declination angle. Calculate the vertical component and the magnitude of the magnetic field vector.
Solution:
The inclination angle can be linked to the horizontal and vertical components of the terrestrial magnetic field: tan I fracB_vB_h Solving for the vertical component yields B_v BvF Bh times tanInc resultBvP Using the same right triangle as before we find that fracB_hB cos I where B is the magnitude of the total magnetic field vector. Solving for B yields B BF fracBhcosInc resultBP Alternatively we can use the pythagorean theorem to calculate the magnitude of the field vector: B BbF sqrtBh^+BvP^ resultBbP
For the magnetic field in Zurich look up the horizontal component the inclination and declination angle. Calculate the vertical component and the magnitude of the magnetic field vector.
Solution:
The inclination angle can be linked to the horizontal and vertical components of the terrestrial magnetic field: tan I fracB_vB_h Solving for the vertical component yields B_v BvF Bh times tanInc resultBvP Using the same right triangle as before we find that fracB_hB cos I where B is the magnitude of the total magnetic field vector. Solving for B yields B BF fracBhcosInc resultBP Alternatively we can use the pythagorean theorem to calculate the magnitude of the field vector: B BbF sqrtBh^+BvP^ resultBbP
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Exercise:
For the magnetic field in Zurich look up the horizontal component the inclination and declination angle. Calculate the vertical component and the magnitude of the magnetic field vector.
Solution:
The inclination angle can be linked to the horizontal and vertical components of the terrestrial magnetic field: tan I fracB_vB_h Solving for the vertical component yields B_v BvF Bh times tanInc resultBvP Using the same right triangle as before we find that fracB_hB cos I where B is the magnitude of the total magnetic field vector. Solving for B yields B BF fracBhcosInc resultBP Alternatively we can use the pythagorean theorem to calculate the magnitude of the field vector: B BbF sqrtBh^+BvP^ resultBbP
For the magnetic field in Zurich look up the horizontal component the inclination and declination angle. Calculate the vertical component and the magnitude of the magnetic field vector.
Solution:
The inclination angle can be linked to the horizontal and vertical components of the terrestrial magnetic field: tan I fracB_vB_h Solving for the vertical component yields B_v BvF Bh times tanInc resultBvP Using the same right triangle as before we find that fracB_hB cos I where B is the magnitude of the total magnetic field vector. Solving for B yields B BF fracBhcosInc resultBP Alternatively we can use the pythagorean theorem to calculate the magnitude of the field vector: B BbF sqrtBh^+BvP^ resultBbP
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