Magnetic field due to a coil wound on a hemsiphere

[palaphys]

Homework Statement

A long insulating wire carrying current $\mathbf{i}$ is wrapped ($\mathbf{N}$ turns) tightly over a hemispherical bowl of radius $\mathbf{R}$ without any gap between consecutive turns up to a height of $\mathbf{\frac{\sqrt{3}R}{2}}$ as shown. The magnetic field at the center is:

Relevant Equations

magnetism equations+integration

this is the figure provided. My approach was to consider the field due to an elemental ring-shape conductor, using the standard formula
$dB = \frac{\mu_0 i r^2 \ dN}{2(x^2 + r^2)^{3/2}}$
where r is the radius of the elemental ring, x is the distance from the center of the hemisphere to that of the ring.

I'm not able to figure out how to calculate dN correctly.
(answer for the question(for magnetic field) is given as :

)

 

[kuruman]

Quantity $dN$ is proportional to what geometrical quantity that you will have to integrate over? Look at the drawing that was provided by the question. What have you tried so far?

 

[palaphys]

Quantity $dN$ is proportional to what geometrical quantity that you will have to integrate over? Look at the drawing that was provided by the question. What have you tried so far?

yes It must be proportional to some quantity. I'm trying to consider its relation with an angle theta, i.e polar angle

 

[kuruman]

yes It must be proportional to some quantity. I'm trying to consider its relation with an angle theta, i.e polar angle

That's a good start. What fraction of the total turns $N$ do you think is within an arc subtending angle $d\theta$?

 

[palaphys]

That's a good start. What fraction of the total turns $N$ do you think is within an arc subtending angle $d\theta$?

I think I will do it this way- for the entire wire covering the arc, the lengthwise turn density is N/(arc length).

here arc length will be $s= R\phi_{\max} = R\pi/3$ based on the geometry of the problem.
so $dN/ds= N/s= \frac{3N}{\pi R}$
where $ds=R(d\theta)$ where theta is the polar angle from +z axis

so far if my ideas are correct, I think I can take it from here.
Is there any other way to find $dN$ ?

 

[kuruman]

Your ideas are correct and you can take it from there. However, since you asked, you can do it more directly by noting that there are $N$ turns distributed over $\frac{\pi}{3}$ so that $$dN=\left(\frac{N}{\pi/3}\right)d\theta=\frac{3N}{\pi}d\theta.$$Since the variable of integration is $\theta$ and not $s$, your approach introduces the extra step $ds=Rd\theta$ but the end result is the same.

 

[palaphys]

However, since you asked, you can do it more directly by noting that there are $N$ turns distributed over $\frac{\pi}{3}$ so that $$dN=\left(\frac{N}{\pi/3}\right)d\theta=\frac{3N}{\pi}d\theta.

Agreed. This is simpler.
Thanks