Magnetic Force As a Vector
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:
An electron with velocity vecv enters a uniform magnetic field vecB. With respect to the axis of a Cartesian coordinate system the two vectors can be written as vecv leftmatrixv_xv_yv_zmatrixright leftmatrixvxMX vyMX vzMXmatrixright unitmegam/s vecB leftmatrixB_xB_yB_zmatrixright leftmatrixBxmX BymX BzmXmatrixright unitmilliT abcliste abc Calculate the components and the magnitude of the magnetic force vector acting on the electron. abc Explain why it is impossible for an electron to experience a force in the direction of the z-axis no matter what direction its velocity vector has. abcliste
Solution:
abcliste abc The force vecF is given by vecF q vecvcrossvecB q left matrixv_y B_z-v_z B_y v_z B_x-v_x B_z v_x B_y-v_y B_x matrix right qtimes left matrixvyMX BzmX-vzMX BymX vzMX BxmX-vxMX BzmX vxMX BymX-vyMX BxmX matrix right unitkilom/stimesT left matrix FfxX FfyX FfzX matrix rightunitfemtonewton approx resultleft matrix FxP- FyP- FzP- matrixright The magnitude of the force vector is absvecFFabsF sqrtleftFfxX^+FfyX^+FfzX^rightunitfemtonewton Fabs approx resultFabsP- abc The magnetic force is always perpicular to both the velocity vecv and the magnetic field vecB. In order to experience a force in the direction of the z-axis the velocity and the magnetic field have to be in the xy-plane. Since vecB has a non-vanishing z component this is obviously not the case. abcliste
An electron with velocity vecv enters a uniform magnetic field vecB. With respect to the axis of a Cartesian coordinate system the two vectors can be written as vecv leftmatrixv_xv_yv_zmatrixright leftmatrixvxMX vyMX vzMXmatrixright unitmegam/s vecB leftmatrixB_xB_yB_zmatrixright leftmatrixBxmX BymX BzmXmatrixright unitmilliT abcliste abc Calculate the components and the magnitude of the magnetic force vector acting on the electron. abc Explain why it is impossible for an electron to experience a force in the direction of the z-axis no matter what direction its velocity vector has. abcliste
Solution:
abcliste abc The force vecF is given by vecF q vecvcrossvecB q left matrixv_y B_z-v_z B_y v_z B_x-v_x B_z v_x B_y-v_y B_x matrix right qtimes left matrixvyMX BzmX-vzMX BymX vzMX BxmX-vxMX BzmX vxMX BymX-vyMX BxmX matrix right unitkilom/stimesT left matrix FfxX FfyX FfzX matrix rightunitfemtonewton approx resultleft matrix FxP- FyP- FzP- matrixright The magnitude of the force vector is absvecFFabsF sqrtleftFfxX^+FfyX^+FfzX^rightunitfemtonewton Fabs approx resultFabsP- abc The magnetic force is always perpicular to both the velocity vecv and the magnetic field vecB. In order to experience a force in the direction of the z-axis the velocity and the magnetic field have to be in the xy-plane. Since vecB has a non-vanishing z component this is obviously not the case. abcliste
Meta Information
Exercise:
An electron with velocity vecv enters a uniform magnetic field vecB. With respect to the axis of a Cartesian coordinate system the two vectors can be written as vecv leftmatrixv_xv_yv_zmatrixright leftmatrixvxMX vyMX vzMXmatrixright unitmegam/s vecB leftmatrixB_xB_yB_zmatrixright leftmatrixBxmX BymX BzmXmatrixright unitmilliT abcliste abc Calculate the components and the magnitude of the magnetic force vector acting on the electron. abc Explain why it is impossible for an electron to experience a force in the direction of the z-axis no matter what direction its velocity vector has. abcliste
Solution:
abcliste abc The force vecF is given by vecF q vecvcrossvecB q left matrixv_y B_z-v_z B_y v_z B_x-v_x B_z v_x B_y-v_y B_x matrix right qtimes left matrixvyMX BzmX-vzMX BymX vzMX BxmX-vxMX BzmX vxMX BymX-vyMX BxmX matrix right unitkilom/stimesT left matrix FfxX FfyX FfzX matrix rightunitfemtonewton approx resultleft matrix FxP- FyP- FzP- matrixright The magnitude of the force vector is absvecFFabsF sqrtleftFfxX^+FfyX^+FfzX^rightunitfemtonewton Fabs approx resultFabsP- abc The magnetic force is always perpicular to both the velocity vecv and the magnetic field vecB. In order to experience a force in the direction of the z-axis the velocity and the magnetic field have to be in the xy-plane. Since vecB has a non-vanishing z component this is obviously not the case. abcliste
An electron with velocity vecv enters a uniform magnetic field vecB. With respect to the axis of a Cartesian coordinate system the two vectors can be written as vecv leftmatrixv_xv_yv_zmatrixright leftmatrixvxMX vyMX vzMXmatrixright unitmegam/s vecB leftmatrixB_xB_yB_zmatrixright leftmatrixBxmX BymX BzmXmatrixright unitmilliT abcliste abc Calculate the components and the magnitude of the magnetic force vector acting on the electron. abc Explain why it is impossible for an electron to experience a force in the direction of the z-axis no matter what direction its velocity vector has. abcliste
Solution:
abcliste abc The force vecF is given by vecF q vecvcrossvecB q left matrixv_y B_z-v_z B_y v_z B_x-v_x B_z v_x B_y-v_y B_x matrix right qtimes left matrixvyMX BzmX-vzMX BymX vzMX BxmX-vxMX BzmX vxMX BymX-vyMX BxmX matrix right unitkilom/stimesT left matrix FfxX FfyX FfzX matrix rightunitfemtonewton approx resultleft matrix FxP- FyP- FzP- matrixright The magnitude of the force vector is absvecFFabsF sqrtleftFfxX^+FfyX^+FfzX^rightunitfemtonewton Fabs approx resultFabsP- abc The magnetic force is always perpicular to both the velocity vecv and the magnetic field vecB. In order to experience a force in the direction of the z-axis the velocity and the magnetic field have to be in the xy-plane. Since vecB has a non-vanishing z component this is obviously not the case. abcliste
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