Magnetic Force on Current
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:
A wire with length LO running from left to right direction of the positive x axis carries a current of IO. It is placed in a homogeneous magnetic field given by the components B_xBxO B_yByO B_zBzO. abcliste abc Calculate the force acting on the wire using the vector version for the magnetic force. abc Determine the angle between the direction of the current and the direction of the magnetic field and calculate the force using the scalar version for the magnetic force. abcliste
Solution:
abcliste abc The force vector is given by vecF I vecLcrossvecB I leftmatrix Lmatrixrightcross leftmatrix B_xB_yB_zmatrixright Ileftmatrix --L B_zL B_y-matrixright I Lleftmatrix - B_z B_ymatrixright The magnitude of the force is therefore F absvecB FF labelsolutiona ItimesLtimessqrtBz^+By^ F approx resultFP- abc The cosine of the angle between vecL and vecB can be found using the dot product: costheta fracvecLvecBabsvecLabsvecB fracL B_xLsqrtB_x^+B_y^+B_z^ fracB_xsqrtB_x^+B_y^+B_z^ Since cos^theta + sin^theta the sine of the angle is sheta sqrt-cos^theta sqrt-fracB_x^B_x^+B_y^+B_z^ sqrtfracB_y^+B_z^B_x^+B_y^+B_z^ Using the scalar version for the magnetic force we thus find F I L B sheta I L sqrtB_x^+B_y^+B_z^ sqrtfracB_y^+B_z^B_x^+B_y^+B_z^ I L sqrtB_y^+B_z^ This is the same expression as refsolutiona so both versions lead to the same result. abcliste
A wire with length LO running from left to right direction of the positive x axis carries a current of IO. It is placed in a homogeneous magnetic field given by the components B_xBxO B_yByO B_zBzO. abcliste abc Calculate the force acting on the wire using the vector version for the magnetic force. abc Determine the angle between the direction of the current and the direction of the magnetic field and calculate the force using the scalar version for the magnetic force. abcliste
Solution:
abcliste abc The force vector is given by vecF I vecLcrossvecB I leftmatrix Lmatrixrightcross leftmatrix B_xB_yB_zmatrixright Ileftmatrix --L B_zL B_y-matrixright I Lleftmatrix - B_z B_ymatrixright The magnitude of the force is therefore F absvecB FF labelsolutiona ItimesLtimessqrtBz^+By^ F approx resultFP- abc The cosine of the angle between vecL and vecB can be found using the dot product: costheta fracvecLvecBabsvecLabsvecB fracL B_xLsqrtB_x^+B_y^+B_z^ fracB_xsqrtB_x^+B_y^+B_z^ Since cos^theta + sin^theta the sine of the angle is sheta sqrt-cos^theta sqrt-fracB_x^B_x^+B_y^+B_z^ sqrtfracB_y^+B_z^B_x^+B_y^+B_z^ Using the scalar version for the magnetic force we thus find F I L B sheta I L sqrtB_x^+B_y^+B_z^ sqrtfracB_y^+B_z^B_x^+B_y^+B_z^ I L sqrtB_y^+B_z^ This is the same expression as refsolutiona so both versions lead to the same result. abcliste
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Exercise:
A wire with length LO running from left to right direction of the positive x axis carries a current of IO. It is placed in a homogeneous magnetic field given by the components B_xBxO B_yByO B_zBzO. abcliste abc Calculate the force acting on the wire using the vector version for the magnetic force. abc Determine the angle between the direction of the current and the direction of the magnetic field and calculate the force using the scalar version for the magnetic force. abcliste
Solution:
abcliste abc The force vector is given by vecF I vecLcrossvecB I leftmatrix Lmatrixrightcross leftmatrix B_xB_yB_zmatrixright Ileftmatrix --L B_zL B_y-matrixright I Lleftmatrix - B_z B_ymatrixright The magnitude of the force is therefore F absvecB FF labelsolutiona ItimesLtimessqrtBz^+By^ F approx resultFP- abc The cosine of the angle between vecL and vecB can be found using the dot product: costheta fracvecLvecBabsvecLabsvecB fracL B_xLsqrtB_x^+B_y^+B_z^ fracB_xsqrtB_x^+B_y^+B_z^ Since cos^theta + sin^theta the sine of the angle is sheta sqrt-cos^theta sqrt-fracB_x^B_x^+B_y^+B_z^ sqrtfracB_y^+B_z^B_x^+B_y^+B_z^ Using the scalar version for the magnetic force we thus find F I L B sheta I L sqrtB_x^+B_y^+B_z^ sqrtfracB_y^+B_z^B_x^+B_y^+B_z^ I L sqrtB_y^+B_z^ This is the same expression as refsolutiona so both versions lead to the same result. abcliste
A wire with length LO running from left to right direction of the positive x axis carries a current of IO. It is placed in a homogeneous magnetic field given by the components B_xBxO B_yByO B_zBzO. abcliste abc Calculate the force acting on the wire using the vector version for the magnetic force. abc Determine the angle between the direction of the current and the direction of the magnetic field and calculate the force using the scalar version for the magnetic force. abcliste
Solution:
abcliste abc The force vector is given by vecF I vecLcrossvecB I leftmatrix Lmatrixrightcross leftmatrix B_xB_yB_zmatrixright Ileftmatrix --L B_zL B_y-matrixright I Lleftmatrix - B_z B_ymatrixright The magnitude of the force is therefore F absvecB FF labelsolutiona ItimesLtimessqrtBz^+By^ F approx resultFP- abc The cosine of the angle between vecL and vecB can be found using the dot product: costheta fracvecLvecBabsvecLabsvecB fracL B_xLsqrtB_x^+B_y^+B_z^ fracB_xsqrtB_x^+B_y^+B_z^ Since cos^theta + sin^theta the sine of the angle is sheta sqrt-cos^theta sqrt-fracB_x^B_x^+B_y^+B_z^ sqrtfracB_y^+B_z^B_x^+B_y^+B_z^ Using the scalar version for the magnetic force we thus find F I L B sheta I L sqrtB_x^+B_y^+B_z^ sqrtfracB_y^+B_z^B_x^+B_y^+B_z^ I L sqrtB_y^+B_z^ This is the same expression as refsolutiona so both versions lead to the same result. abcliste
Contained in these collections:
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Magnetic Forces by by
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Force Vectors by by