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