Can not understand how to divide the components of the vector.

In summary: This is helpful because now we know that the force on the straight segment is cosθF1, and the force on the arc is -sinθF2.
  • #1
arierreF
79
0

Homework Statement



Consider a closed semi-circular loop lying in the xy plane carrying a current I in the
counterclockwise direction, as shown in Figure attached.

2.png


A uniform magnetic field pointing in the +y direction is applied. Find the magnetic force acting on the straight segment and the semicircular arc.



Homework Equations



The force in xx axis [itex]\vec{F1}[/itex] is easy to see that it has the normal direction [itex]\hat{i}[/itex]

so that force :

[itex] \vec{F} = I 2R\hat{i}\times \vec{B} \hat{j}= 2IRB \hat{k} [/itex]

where [itex]\hat{k}[/itex] is directed out the page.

Now the force along the arc.


The solution says:

To evaluate [itex]\vec{F2}[/itex] , we first note that the differential length element [itex]d\vec{s}[/itex] on the semicircle can be written as:

[itex]d\vec{s} = ds\hat{\theta } = IRd\theta (-sin \theta \hat{i} + cos \theta \hat{j})[/itex]


I know that [itex]s = R \theta[/itex], but i don't know where [itex]-sin \theta \hat{i} + cos \theta \hat{j} [/itex] come from.



Some tips ??
 
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  • #2
The vector ##\vec{ds}## at a point on the circular arc is tangent to the circular arc and therefore perpendicular to the radius. (The diagram doesn't really show it as perpendicular, but that's because they've drawn a fairly large ##\vec{ds}## vector. It is actually infinitesimally small.) ##\theta## is the angle that the radius line makes to the horizontal. Can you deduce the angle that ##\vec{ds}## makes to the horizontal or the vertical?

Should your expression [itex]d\vec{s} = ds\hat{\theta } = IRd\theta (-sin \theta \hat{i} + cos \theta \hat{j})[/itex] have the current I in it?
 
  • #3
TSny said:
The vector ##\vec{ds}## at a point on the circular arc is tangent to the circular arc and therefore perpendicular to the radius. (The diagram doesn't really show it as perpendicular, but that's because they've drawn a fairly large ##\vec{ds}## vector. It is actually infinitesimally small.) ##\theta## is the angle that the radius line makes to the horizontal. Can you deduce the angle that ##\vec{ds}## makes to the horizontal or the vertical?

Should your expression [itex]d\vec{s} = ds\hat{\theta } = IRd\theta (-sin \theta \hat{i} + cos \theta \hat{j})[/itex] have the current I in it?

Actually it hasn't the current in the expression (my mistake)

At first glance it seems that [itex]d\vec{s}[/itex] makes a [itex]\theta [/itex] angle with the horizontal. But that doesn't make sense because the component of the horizontal would be [itex](-cos \theta \hat{i}[/itex])
 
Last edited:
  • #4
arierreF said:

Homework Statement



Consider a closed semi-circular loop lying in the xy plane carrying a current I in the
counterclockwise direction, as shown in Figure attached.

View attachment 54474

A uniform magnetic field pointing in the +y direction is applied. Find the magnetic force acting on the straight segment and the semicircular arc.



Homework Equations



The force in xx axis [itex]\vec{F1}[/itex] is easy to see that it has the normal direction [itex]\hat{i}[/itex]

so that force :

[itex] \vec{F} = I 2R\hat{i}\times \vec{B} \hat{j}= 2IRB \hat{k} [/itex]

where [itex]\hat{k}[/itex] is directed out the page.

Now the force along the arc.


The solution says:

To evaluate [itex]\vec{F2}[/itex] , we first note that the differential length element [itex]d\vec{s}[/itex] on the semicircle can be written as:

[itex]d\vec{s} = ds\hat{\theta } = IRd\theta (-sin \theta \hat{i} + cos \theta \hat{j})[/itex]


I know that [itex]s = R \theta[/itex], but i don't know where [itex]-sin \theta \hat{i} + cos \theta \hat{j} [/itex] come from.



Some tips ??

[itex]-sin \theta \hat{i} + cos \theta \hat{j} [/itex] is a unit vector pointing in the θ direction, expressed in terms of the unit vectors for the x-y cartesian coordinate system. The differential position vector ds is pointing in the θ direction. Its magnitude is Rdθ.
 
  • #5
But the problem is that i don't understand why it is not [itex]-cos \theta \hat{i} + sin \theta \hat{j}[/itex] because the vector [itex]\vec{ds}[/itex] makes a angle [itex]theta[/itex] with the horizontal.
 
  • #6
hum, maybe now it is correct?

p5.png


The angle that is does with the horizontal is [itex]90 - \theta[/itex] so it is [itex]cos(90 - \theta) = sin(\theta) [/itex]
 
  • #7
arierreF said:
hum, maybe now it is correct?

View attachment 54476

The angle that is does with the horizontal is [itex]90 - \theta[/itex] so it is [itex]cos(90 - \theta) = sin(\theta) [/itex]

Yes, that's right.
 
  • #8
Now it is understood!

Thanks again!
 
  • #9
A trick that helps sometimes (or at least allows you to check your answer) is to consider the unit vector at the limits θ=0 and at θ=π/2. If you are considering the unit vector in the θ direction, then at θ=0, the θ-direction unit vector is +j, and at θ=π/2, the θ-direction unit vector is -i. We also know that at θ=0, cosθ=1, and sinθ=0, while at θ=π/2, cosθ=0, and sinθ=1. Therefore, the coefficient of j must be cosθ, and the coefficient of i must be -sinθ.
 

Related to Can not understand how to divide the components of the vector.

1. How do I divide the components of a vector?

To divide the components of a vector, you first need to understand the concept of vector multiplication. A vector can be multiplied by a scalar value, which is a number, to change its magnitude. This is known as scalar multiplication. To divide a vector, you would use the inverse of scalar multiplication, which is known as scalar division. This means dividing each component of the vector by the scalar value.

2. Can you give an example of dividing the components of a vector?

Yes, let's say we have a vector V = (2, 4, 6) and we want to divide it by a scalar value of 2. To do this, we would divide each component of the vector by 2, giving us a new vector V' = (1, 2, 3). This means that each component of the original vector has been halved.

3. Why is it important to understand how to divide the components of a vector?

Understanding how to divide the components of a vector is important because vectors are used in many fields of science, such as physics and engineering. Being able to manipulate vectors, including dividing them, allows for more accurate calculations and analysis of data. It is also a fundamental concept in linear algebra, which is used in many scientific applications.

4. Are there any rules or limitations to dividing vector components?

Yes, there are a few rules to keep in mind when dividing vector components. One rule is that you cannot divide a vector by 0, as division by 0 is undefined. Additionally, the order of the components in a vector matters. Dividing a vector by a scalar value will result in a different vector than dividing the scalar by the vector. Lastly, the number of components in the vector must match the number of components in the scalar value.

5. How can I practice and improve my skills in dividing vector components?

A good way to practice and improve your skills in dividing vector components is to work through practice problems and exercises. There are many online resources and textbooks available that offer these types of problems. Additionally, you can also try creating your own problems and solutions to further solidify your understanding of the concept.

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