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

[arierreF]

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.

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 ??

 

[TSny]

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?

 

[arierreF]

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])

 

[Chestermiller]

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.

 

[arierreF]

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.

 

[arierreF]

hum, maybe now it is correct?

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

 

[TSny]

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.

 

[arierreF]

Now it is understood!

Thanks again!

 

[Chestermiller]

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.