Help Understanding Biot-Savart Law Issue

[difftoint]

I've been looking at my electromagnetism lecture notes I have been given and am trying to make sense of a step made in parameterising a current loop.

I am really confused about the part where dl is first state. I have no idea how the 0, sine and cosine got there. I suspect that it's something to do with arc lengths, however. Although I'm not 100% sure.

In short, I need help, desperately.

 

[diazona]

Well for starters, consider a circle of radius r drawn on a two-dimensional graph, centered at the origin. Do you understand that the circle can be described as the set of points
(r\cos\theta, r\sin\theta)
for 0 \leq \theta \leq 2\pi?

 

[difftoint]

Well for starters, consider a circle of radius r drawn on a two-dimensional graph, centered at the origin. Do you understand that the circle can be described as the set of points
(r\cos\theta, r\sin\theta)
for 0 \leq \theta \leq 2\pi?

I understand that part with no difficulties. It explains the first step in which point r_c is denoted.

It's just the second stage that I'm not seeing clearly. It looks like a(0, -sin\theta, cos\theta) is arrived at by differentiation of r_c, but I cannot make the connection between doing that and finding dl.

 

[qbert]

there are a couple of ways to look at this.

the easiest is probably that dl is a tangent vector to the
path that the current takes. and if we have a function
r(t) = (x(t), y(t), z(t)) then the tangent vector is given
by r'(t) = (x'(t), y'(t), z'(t)).


more geometrically we can guess what the tangent vector
should look like. it's length should be the "infinitessimal" arc
so a d\phi and should be pointing in the direction
the current is going. a unit vector pointing to the spot on the
circle is (0, cos, sin), so a unit vector tangent to the circle is
+/- (0, -sin, cos) where we pick the sign to match the direction
the current is going.

how did i know to pick (0, -sin, cos)? well here's an argument.
for points on the circle r^2 = a^2, so we differentiate to see
r(\phi) \cdot \frac{dr}{d\phi} = 0 so that r is perpendicular to
\frac{dr}{d\phi}. then we only need to guess something that's
perpendicular...

 

[difftoint]

there are a couple of ways to look at this.

the easiest is probably that dl is a tangent vector to the
path that the current takes. and if we have a function
r(t) = (x(t), y(t), z(t)) then the tangent vector is given
by r'(t) = (x'(t), y'(t), z'(t)).


more geometrically we can guess what the tangent vector
should look like. it's length should be the "infinitessimal" arc
so a d\phi and should be pointing in the direction
the current is going. a unit vector pointing to the spot on the
circle is (0, cos, sin), so a unit vector tangent to the circle is
+/- (0, -sin, cos) where we pick the sign to match the direction
the current is going.

how did i know to pick (0, -sin, cos)? well here's an argument.
for points on the circle r^2 = a^2, so we differentiate to see
r(\phi) \cdot \frac{dr}{d\phi} = 0 so that r is perpendicular to
\frac{dr}{d\phi}. then we only need to guess something that's
perpendicular...

Thank you so much for the help. Your reply explains the solution really well.