How Does the Biot-Savart Law Apply to a Wire Segment with a Circular Arc?

[22990atinesh]

Homework Statement

Calculate the magnetic field at point O for the current-carrying wire segment shown in Figure. The wire consists of two straight portions and a circular arc of radius R, which subtends an angle Ɵ. The arrowheads on the wire indicate the direction of the current(Biot–Savart Law).

2. The attempt at a solution

$dB = \frac {μ I}{4π} \frac {ds}{R^2}$ - I

$B = \frac {μ I}{4π R^2} \int ds = \frac {μ I}{4π R^2} s = \frac {μ I}{4π R} Θ$

Since s=RΘ

I'm little bit confused in the calcualtion of ds in equation I.
$|\vec {ds} \times \hat r| = |\vec {ds}|.1.sin 90^o = ds$
Does ds here represent the magnitude of the vector perpendicular to the vector $\vec {ds}$ and $\hat r$ or it represents the magnitude of orginal vector $\vec {ds}$.

 

[dauto]

ds is the magnitude of the vector ds what else could it be?

 

[22990atinesh]

ds is the magnitude of the vector ds what else could it be?

I know ds is magnitude of $\vec{ds}$. My doubt is What does it represents. Does ds here represent the magnitude of the vector perpendicular to the vector $\vec{ds}$ and $\hat{r}$ or it represents the magnitude of orginal vector $\vec{ds}$.

 

[AGNuke]

ds is the magnitude of the original vector $\vec{ds}$ and the direction of $\vec {ds} \times \hat r$ is the direction of $d\vec{B}$ due to the elemental current carrying wire.

 

[22990atinesh]

ds is the magnitude of the original vector $\vec{ds}$ and the direction of $\vec {ds} \times \hat r$ is the direction of $d\vec{B}$ due to the elemental current carrying wire.

Thanx I get it