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 \vec{F1} is easy to see that it has the normal direction \hat{i}

so that force :

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

where \hat{k} is directed out the page.

Now the force along the arc.


The solution says:

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

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


I know that s = R \theta, but i don't know where -sin \theta \hat{i} + cos \theta \hat{j} 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 d\vec{s} = ds\hat{\theta } = IRd\theta (-sin \theta \hat{i} + cos \theta \hat{j}) 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 d\vec{s} = ds\hat{\theta } = IRd\theta (-sin \theta \hat{i} + cos \theta \hat{j}) have the current I in it?

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

At first glance it seems that d\vec{s} makes a \theta angle with the horizontal. But that doesn't make sense because the component of the horizontal would be (-cos \theta \hat{i})

 

[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 \vec{F1} is easy to see that it has the normal direction \hat{i}

so that force :

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

where \hat{k} is directed out the page.

Now the force along the arc.


The solution says:

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

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


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



Some tips ??

-sin \theta \hat{i} + cos \theta \hat{j} 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 -cos \theta \hat{i} + sin \theta \hat{j} because the vector \vec{ds} makes a angle theta with the horizontal.

 

[arierreF]

hum, maybe now it is correct?

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

 

[TSny]

hum, maybe now it is correct?

View attachment 54476

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

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.