Helmholtz Coil
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
\(\LaTeX\)
No explanation / solution video to this exercise has yet been created.
Visit our YouTube-Channel to see solutions to other exercises.
Don't forget to subscribe to our channel, like the videos and leave comments!
Visit our YouTube-Channel to see solutions to other exercises.
Don't forget to subscribe to our channel, like the videos and leave comments!
Exercise:
The magnetic field along the axis of a Helmholtz coil with radius rO and NO turns per coil is measured to be BHO. abcliste abc Calculate the current through the coils. abc The distance between the coils is increased to dO while keeping the current constant. Calculate the magnetic field in the middle between the two coils. abcliste
Solution:
abcliste abc The magnetic field in a Helmholtz coil is given by sscBH mu_leftfracright^fracfracN Ir Solving for the current I leads to I leftfracright^fracfracsscBH rmu_ N leftfracright^fractimesfracBHtimes rncmuotimes N I approx resultIP abc The distance corresponds to twice the radius d r. For a single coil the magnetic field at a distance r from its centre can be calculated as B_r fracmu_fracN I r^leftr^+r^right^frac fracmu_fracN I r^left r^right^frac fracmu_fracN I r^^frac r^ fracmu_ ^fracfracN Ir fracmu_ leftfracright^fracfracN Ir For both coils together the magnetic field at the centre is given by sscBtot BF ncmuotimes leftfracright^fractimesfracNtimes Ir B approx resultBP- abcliste
The magnetic field along the axis of a Helmholtz coil with radius rO and NO turns per coil is measured to be BHO. abcliste abc Calculate the current through the coils. abc The distance between the coils is increased to dO while keeping the current constant. Calculate the magnetic field in the middle between the two coils. abcliste
Solution:
abcliste abc The magnetic field in a Helmholtz coil is given by sscBH mu_leftfracright^fracfracN Ir Solving for the current I leads to I leftfracright^fracfracsscBH rmu_ N leftfracright^fractimesfracBHtimes rncmuotimes N I approx resultIP abc The distance corresponds to twice the radius d r. For a single coil the magnetic field at a distance r from its centre can be calculated as B_r fracmu_fracN I r^leftr^+r^right^frac fracmu_fracN I r^left r^right^frac fracmu_fracN I r^^frac r^ fracmu_ ^fracfracN Ir fracmu_ leftfracright^fracfracN Ir For both coils together the magnetic field at the centre is given by sscBtot BF ncmuotimes leftfracright^fractimesfracNtimes Ir B approx resultBP- abcliste
Meta Information
Exercise:
The magnetic field along the axis of a Helmholtz coil with radius rO and NO turns per coil is measured to be BHO. abcliste abc Calculate the current through the coils. abc The distance between the coils is increased to dO while keeping the current constant. Calculate the magnetic field in the middle between the two coils. abcliste
Solution:
abcliste abc The magnetic field in a Helmholtz coil is given by sscBH mu_leftfracright^fracfracN Ir Solving for the current I leads to I leftfracright^fracfracsscBH rmu_ N leftfracright^fractimesfracBHtimes rncmuotimes N I approx resultIP abc The distance corresponds to twice the radius d r. For a single coil the magnetic field at a distance r from its centre can be calculated as B_r fracmu_fracN I r^leftr^+r^right^frac fracmu_fracN I r^left r^right^frac fracmu_fracN I r^^frac r^ fracmu_ ^fracfracN Ir fracmu_ leftfracright^fracfracN Ir For both coils together the magnetic field at the centre is given by sscBtot BF ncmuotimes leftfracright^fractimesfracNtimes Ir B approx resultBP- abcliste
The magnetic field along the axis of a Helmholtz coil with radius rO and NO turns per coil is measured to be BHO. abcliste abc Calculate the current through the coils. abc The distance between the coils is increased to dO while keeping the current constant. Calculate the magnetic field in the middle between the two coils. abcliste
Solution:
abcliste abc The magnetic field in a Helmholtz coil is given by sscBH mu_leftfracright^fracfracN Ir Solving for the current I leads to I leftfracright^fracfracsscBH rmu_ N leftfracright^fractimesfracBHtimes rncmuotimes N I approx resultIP abc The distance corresponds to twice the radius d r. For a single coil the magnetic field at a distance r from its centre can be calculated as B_r fracmu_fracN I r^leftr^+r^right^frac fracmu_fracN I r^left r^right^frac fracmu_fracN I r^^frac r^ fracmu_ ^fracfracN Ir fracmu_ leftfracright^fracfracN Ir For both coils together the magnetic field at the centre is given by sscBtot BF ncmuotimes leftfracright^fractimesfracNtimes Ir B approx resultBP- abcliste
Contained in these collections: