Moving Rod
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:
A rod with length LO and diameter dO is oriented with its length parallel to the east-west direction. It is moved in the earth's magnetic field in Zurich according to the figure top view. The velocity in each of the cases a b and c is vO. Calculate the emf induced in the wire and determine its polarity. center includegraphicswidthtextwidth#image_path:moving-rod# center
Solution:
abcliste abc The relevant component of the earth's magnetic field is the vertical component B_v which can be expressed in terms of the horizontal component B_h and the inclination angle I: B_v B_h tan I The induced emf is therefore mathcalE_a v_a B_v L VaF vtimes BhtimestanIdegtimes L Va approx resultVaP- The vertical component of the magnetic field in Zurich pos downwards o the sheet. Using the left-hand-rule we find that the electrons experience a force to the right so the negative pole of the induced emf is at the right of the rod. center includegraphicswidth.mm#image_path:moving-rod-solution-a-# center abc Again the vertical component is the relevant component. The difference to a is that in b the emf is induced across the diameter of the rod. mathcalE_b v_b B_v d VbF vtimes BhtimestanIdegtimes d Vb approx resultVbP- The polarity can be found using the left-hand-rule. center includegraphicswidth.mm#image_path:moving-rod-solution-b-# center abc Since the rod moves in the vertical direction we have to take o account the horizontal component of the magnetic field: mathcalE_c VcF vtimes Bhtimes L Vc approx resultVcP- The polarity once again follows from the left-hand-rule. center includegraphicswidth.mm#image_path:moving-rod-solution-c# center abcliste
A rod with length LO and diameter dO is oriented with its length parallel to the east-west direction. It is moved in the earth's magnetic field in Zurich according to the figure top view. The velocity in each of the cases a b and c is vO. Calculate the emf induced in the wire and determine its polarity. center includegraphicswidthtextwidth#image_path:moving-rod# center
Solution:
abcliste abc The relevant component of the earth's magnetic field is the vertical component B_v which can be expressed in terms of the horizontal component B_h and the inclination angle I: B_v B_h tan I The induced emf is therefore mathcalE_a v_a B_v L VaF vtimes BhtimestanIdegtimes L Va approx resultVaP- The vertical component of the magnetic field in Zurich pos downwards o the sheet. Using the left-hand-rule we find that the electrons experience a force to the right so the negative pole of the induced emf is at the right of the rod. center includegraphicswidth.mm#image_path:moving-rod-solution-a-# center abc Again the vertical component is the relevant component. The difference to a is that in b the emf is induced across the diameter of the rod. mathcalE_b v_b B_v d VbF vtimes BhtimestanIdegtimes d Vb approx resultVbP- The polarity can be found using the left-hand-rule. center includegraphicswidth.mm#image_path:moving-rod-solution-b-# center abc Since the rod moves in the vertical direction we have to take o account the horizontal component of the magnetic field: mathcalE_c VcF vtimes Bhtimes L Vc approx resultVcP- The polarity once again follows from the left-hand-rule. center includegraphicswidth.mm#image_path:moving-rod-solution-c# center abcliste
Meta Information
Exercise:
A rod with length LO and diameter dO is oriented with its length parallel to the east-west direction. It is moved in the earth's magnetic field in Zurich according to the figure top view. The velocity in each of the cases a b and c is vO. Calculate the emf induced in the wire and determine its polarity. center includegraphicswidthtextwidth#image_path:moving-rod# center
Solution:
abcliste abc The relevant component of the earth's magnetic field is the vertical component B_v which can be expressed in terms of the horizontal component B_h and the inclination angle I: B_v B_h tan I The induced emf is therefore mathcalE_a v_a B_v L VaF vtimes BhtimestanIdegtimes L Va approx resultVaP- The vertical component of the magnetic field in Zurich pos downwards o the sheet. Using the left-hand-rule we find that the electrons experience a force to the right so the negative pole of the induced emf is at the right of the rod. center includegraphicswidth.mm#image_path:moving-rod-solution-a-# center abc Again the vertical component is the relevant component. The difference to a is that in b the emf is induced across the diameter of the rod. mathcalE_b v_b B_v d VbF vtimes BhtimestanIdegtimes d Vb approx resultVbP- The polarity can be found using the left-hand-rule. center includegraphicswidth.mm#image_path:moving-rod-solution-b-# center abc Since the rod moves in the vertical direction we have to take o account the horizontal component of the magnetic field: mathcalE_c VcF vtimes Bhtimes L Vc approx resultVcP- The polarity once again follows from the left-hand-rule. center includegraphicswidth.mm#image_path:moving-rod-solution-c# center abcliste
A rod with length LO and diameter dO is oriented with its length parallel to the east-west direction. It is moved in the earth's magnetic field in Zurich according to the figure top view. The velocity in each of the cases a b and c is vO. Calculate the emf induced in the wire and determine its polarity. center includegraphicswidthtextwidth#image_path:moving-rod# center
Solution:
abcliste abc The relevant component of the earth's magnetic field is the vertical component B_v which can be expressed in terms of the horizontal component B_h and the inclination angle I: B_v B_h tan I The induced emf is therefore mathcalE_a v_a B_v L VaF vtimes BhtimestanIdegtimes L Va approx resultVaP- The vertical component of the magnetic field in Zurich pos downwards o the sheet. Using the left-hand-rule we find that the electrons experience a force to the right so the negative pole of the induced emf is at the right of the rod. center includegraphicswidth.mm#image_path:moving-rod-solution-a-# center abc Again the vertical component is the relevant component. The difference to a is that in b the emf is induced across the diameter of the rod. mathcalE_b v_b B_v d VbF vtimes BhtimestanIdegtimes d Vb approx resultVbP- The polarity can be found using the left-hand-rule. center includegraphicswidth.mm#image_path:moving-rod-solution-b-# center abc Since the rod moves in the vertical direction we have to take o account the horizontal component of the magnetic field: mathcalE_c VcF vtimes Bhtimes L Vc approx resultVcP- The polarity once again follows from the left-hand-rule. center includegraphicswidth.mm#image_path:moving-rod-solution-c# center abcliste
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