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

## 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 ??

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TSny
Homework Helper
Gold Member
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?

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 $(-cos \theta \hat{i}$)

Last edited:
Chestermiller
Mentor

## 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.

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.

TSny
Homework Helper
Gold Member
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.

Now it is understood!

Thanks again!

Chestermiller
Mentor
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.