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Yesterday I tried to find an expression to predict the behaviour of Magnetic Field due a circular loop at any point in the space, in cylindrical co-ordinates. At the end, I am stuck with some integrals.

Diagram:

enter image description here


Process:

Initially I found out the position vector of a fixed point $\displaystyle P(\rho_0, \phi_0 , z_0)$ : $\displaystyle \vec{r_0} = \rho_0 \hat{\rho_0} + z_0 \hat{z}$

And a general point on loop $\displaystyle O(\rho, \phi , 0)$ : $\displaystyle \vec{r} = \rho \hat{\rho} \implies \displaystyle \vec{r^{'}} = \rho_0 \hat{\rho_0} + z_0 \hat{z} - \rho \hat{\rho} $

As the angle between $\displaystyle \hat{\rho} \ \text{and} \ \hat{\rho_0} $ is $ (\phi - \phi_0) $ we can derive: $\displaystyle \hat{\rho} = \cos (\phi - \phi_0 ) \hat{\rho_0} + \sin (\phi - \phi_0 ) \hat{\phi_0} $.

Now, the general line element in cylindrical co-ordinates is : $\displaystyle \vec{\mathrm{d}l} = \mathrm{d} \rho \ \hat{\rho} + \rho \mathrm{d}\phi \ \hat{\phi} + \mathrm{d} z \ \hat{z} \ $ but here,

$$ \displaystyle \ \text{line element:} \ \vec{\mathrm{d}l} = \rho \mathrm{d}\phi \ \hat{\phi} $$


Biot-Savart Law: $ \displaystyle \vec{B} = \frac{\mu_0}{4\pi} \int_{c} \frac{i \vec{\mathrm{d}l} \times \vec{r^{'}}}{||\vec{r^{'}}||^3}$

Hence, first we need to find the cross product $\displaystyle \vec{\mathrm{d}l} \times \vec{r^{'}} = \rho \mathrm{d}\phi \ \hat{\phi} \times \vec{r^{'}} $

$$\displaystyle \hat{\phi} \times \vec{r^{'}} = z_0 \cos (\phi - \phi_0) \hat{\rho_0} + z_0 \sin (\phi - \phi_0 ) \hat{\phi_0} - (\rho + \rho_0 \cos (\phi - \phi_0)) \hat{z}$$

And $ \displaystyle ||\vec{r^{'}}|| = \left[ z^{2}_{0} + \rho_0^{2} + \rho^{2} - 2\rho_0 \rho \cos (\phi - \phi_0) \right]^{1/2} $

Thus our Integral becomes:

$$ \displaystyle \vec{B} = \frac{\mu_0 i \rho }{4\pi} \int_{0}^{2\pi} \frac{z_0 \cos (\phi - \phi_0) \hat{\rho_0} + z_0 \sin (\phi - \phi_0 ) \hat{\phi_0} - (\rho + \rho_0 \cos (\phi - \phi_0)) \hat{z}}{\left[ z^{2}_{0} + \rho_0^{2} + \rho^{2} - 2\rho_0 \rho \cos (\phi - \phi_0) \right]^{3/2} } \mathrm{d} \phi $$

The coefficient of $\displaystyle \hat{\phi_0} $ is integrable and the definite integral here, results to be zero. This have to be true according to laws of electromagnetism (i.e no magnetic field parallel to the current element).

Hence our Magnetic field becomes:

$$ \displaystyle \vec{B} = \frac{\mu_0 i \rho z_0 }{4\pi} \int_{0}^{2\pi} \frac{\cos (\phi - \phi_0)}{\left[ z^{2}_{0} + \rho_0^{2} + \rho^{2} - 2\rho_0 \rho \cos (\phi - \phi_0) \right]^{3/2} } \mathrm{d} \phi \ \hat{\rho_0} \\ - \frac{\mu_0 i \rho }{4\pi} \int_{0}^{2\pi} \frac{\rho + \rho_0 \cos (\phi - \phi_0)}{\left[ z^{2}_{0} + \rho_0^{2} + \rho^{2} - 2\rho_0 \rho \cos (\phi - \phi_0) \right]^{3/2} } \mathrm{d} \phi \ \hat{z} $$


EDIT:

I have used the Ellipltical integrals (as suggested by @Maxim) to simplify the expression as:

$$ \displaystyle \vec{B} = \frac{\mu_0 i \sqrt{a} }{2^{2.5} \pi \rho_0^{1.5} \sqrt{\rho} \sqrt{1-a}} (z_0-z) \left [ \frac{1}{1+a} \mathrm{E}\left(\pi, \frac{2a}{1-a} \right) - \mathrm{F}\left( \pi,\frac{2a}{1-a} \right)\right ] \hat{\rho_0} - \frac{\mu_0 i \sqrt{a}}{2^{2.5} \pi \rho_0^{1.5} \sqrt{\rho} \sqrt{1-a}} \left [ \left(\frac{\rho_0}{a+1} + \rho \right) \mathrm{E}\left(\pi, \frac{2a}{1-a} \right) - \rho_0 \mathrm{F}\left( \pi,\frac{2a}{1-a} \right)\right ] \hat{z}\\ $$

Where $\displaystyle a = \frac{2\rho \rho_0}{(z_0-z)^2 + \rho^2 + \rho_0^2}$ and the position of the ring is $(\rho,\phi,z).$

$\mathrm{F}$ is elliptical integral of first kind and $\mathrm{E}$ is of second kind.


Thus, my intial stage is well simplified. But the reason behind this complex calculation was that, I want to complute the Magnetic field due to a solenoid at any point in sapce.

If we assume the number density of the coil to be $n$ and length $z_f - z_i,$ then in a small region of thickness $\mathrm{d}z$, there will be $n\mathrm{d}z$ rings, each with nearly $\vec{B}$ as their magnetic field.

Thus $$\displaystyle \vec{B}_{\text{solenoid}} = \int_{z_i}^{z_f} n\vec{B}\mathrm{d}z $$

Hence it is equivalent to compute :

$$ \displaystyle \vec{B}_{\text{solenoid}} = \frac{\mu_0 n i }{2^{2.5} \pi \rho_0^{1.5} \sqrt{\rho}} \int_{z_i}^{z_f} \sqrt{\frac{a}{1-a}} \left( (z_0-z) \left [ \frac{1}{1+a} \mathrm{E}\left(\pi, \frac{2a}{1-a} \right) - \mathrm{F}\left( \pi,\frac{2a}{1-a} \right)\right ] \hat{\rho_0} - \left [ \left(\frac{\rho_0}{a+1} + \rho \right) \mathrm{E}\left(\pi, \frac{2a}{1-a} \right) - \rho_0 \mathrm{F}\left( \pi,\frac{2a}{1-a} \right)\right ] \hat{z} \right) \mathrm{d}z $$


Thus I need help in solving,

$\displaystyle \int_{z_i}^{z^f} \frac {(z_0-z) ( ( z-z_0)^2 +\rho^2 + \rho_0^2 ) }{\sqrt{( z-z_0)^2 + (\rho - \rho_0)^2} (( z-z_0)^2 + (\rho + \rho_0)^2) } \mathrm{E} \left( \pi, \frac{4\rho \rho_0}{ (z-z_0)^2 + (\rho - \rho_0)^2 } \right) \mathrm{d}z $

$\displaystyle \int_{z_i}^{z^f} \frac {1 }{\sqrt{( z-z_0)^2 + (\rho - \rho_0)^2} } \mathrm{E} \left( \pi, \frac{4\rho \rho_0}{ (z-z_0)^2 + (\rho - \rho_0)^2 } \right) \mathrm{d}z $

$\displaystyle \int_{z_i}^{z^f} \frac {(z_0-z) }{\sqrt{( z-z_0)^2 + (\rho - \rho_0)^2} } \mathrm{F} \left( \pi, \frac{4\rho \rho_0}{ (z-z_0)^2 + (\rho - \rho_0)^2 } \right) \mathrm{d}z $

$\displaystyle \int_{z_i}^{z^f} \frac {1 }{\sqrt{( z-z_0)^2 + (\rho - \rho_0)^2} } \mathrm{F} \left( \pi, \frac{4\rho \rho_0}{ (z-z_0)^2 + (\rho - \rho_0)^2 } \right) \mathrm{d}z $

in order to solve for my expression


Question: The above formula should give Magnetic field of solenoid at any point in space. Do we have any closed form for the above integral ? Or it's taylor series that could simplify the expression.


Progress:

Edit $(30/03/2022) :$ I have got a lead, we can break $\mathrm{E}(\pi,k)$ into complete elliptical integral as

$$\displaystyle \mathrm{E}(\pi,k) = \mathrm{E}(k) + \mathrm{E}\left(\sqrt{\frac{k^2}{1-k^2}}\right)$$

and using the expansion we will get:

$$\displaystyle \mathrm{E}(\pi,k) = \pi - \frac{\pi}{2} \sum_{n=0}^{\infty} \frac{1}{16^n (2n-1)} {\binom{2n}{n}}^2 \left( k^{2n} + {\left[\frac{k^2}{1-k^2} \right]}^n \right)$$

This does simplify our integrand, but it is still vey complicated to sort out.

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  • $\begingroup$ integral-calculator.com Always a good idea to check this handy site once! $\endgroup$ Commented Mar 23, 2022 at 9:29
  • $\begingroup$ @RaadShaikh It is not useful, as it is long to solve and shows TIME OUT. $\endgroup$ Commented Mar 23, 2022 at 11:04
  • $\begingroup$ @RaadShaikh Thank you for sharing the link, it is handy as I was able to get numerical approximations at certain point. $\endgroup$ Commented Mar 23, 2022 at 11:25
  • $\begingroup$ @RaadShaikh The only issue is, it can't solve the integral with variables. I was looking for solution in variables. $\endgroup$ Commented Mar 23, 2022 at 11:27
  • $\begingroup$ Considering only $\phi$ is the variable of integration, we could consider a 'stripped-down' version of this integral by simplifying the constants and splitting it into terms like $\cos\phi*(1-\cos\phi)^{-3/2}, (1-\cos\phi)^{-3/2}$. Both of these have analytic (if complicated) solutions that you could check on wolframalpha.com $\endgroup$ Commented Mar 23, 2022 at 12:31

2 Answers 2

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Suppose we have a finite continuous solenoid of radius $\rho$ and axial length $L$ where $I$ is the current carried by the wire and $n$ is the number of turns of wire per unit length. For convenience, we place the center of the solenoid at the origin and orient its axis to align with the $z$-axis.

The magnetic field $\vec{B}$ at the point $\vec{r}_{0}$ due to the solenoid is given by the iterated integral

$$\vec{B}{\left(\vec{r}_{0}\right)}=n\int_{-\frac{L}{2}}^{\frac{L}{2}}\mathrm{d}z\,\frac{\mu_{0}I}{4\pi}\int_{0}^{2\pi}\mathrm{d}\phi\,\frac{\frac{\partial\vec{r}}{\partial\phi}\times\left(\vec{r}_{0}-\vec{r}\right)}{\|\vec{r}_{0}-\vec{r}\|^{3}},$$

where $\vec{r}$ is the position vector for a point with cylindrical coordinates $\left(\rho,\phi,z\right)$ on the surface of the solenoid:

$$\vec{r}=\rho\cos{\left(\phi\right)}\,\hat{x}+\rho\sin{\left(\phi\right)}\,\hat{y}+z\hat{z}.$$

Similarly, the position vector $\vec{r}_{0}$ of a point with cylindrical coordinates $\left(\rho_{0},\phi_{0},z_{0}\right)$ can be written as

$$\vec{r}_{0}=\rho_{0}\cos{\left(\phi_{0}\right)}\,\hat{x}+\rho_{0}\sin{\left(\phi_{0}\right)}\,\hat{y}+z_{0}\hat{z}.$$

A closed-form expression for $\vec{B}{\left(\vec{r}_{0}\right)}$ in terms of complete elliptic integrals is derived below.

For reference, the complete elliptic integrals are defined here as follows:

$$K{\left(\kappa\right)}=\int_{0}^{\frac{\pi}{2}}\mathrm{d}\varphi\,\frac{1}{\sqrt{1-\kappa^{2}\sin^{2}{\left(\varphi\right)}}};~~~\small{\kappa\in\left(-1,1\right)},$$

$$E{\left(\kappa\right)}=\int_{0}^{\frac{\pi}{2}}\mathrm{d}\varphi\,\sqrt{1-\kappa^{2}\sin^{2}{\left(\varphi\right)}};~~~\small{\kappa\in\left[-1,1\right]},$$

$$\Pi{\left(\nu,\kappa\right)}=\int_{0}^{\frac{\pi}{2}}\mathrm{d}\varphi\,\frac{1}{\left[1-\nu\sin^{2}{\left(\varphi\right)}\right]\sqrt{1-\kappa^{2}\sin^{2}{\left(\varphi\right)}}};~~~\small{\kappa\in\left(-1,1\right)\land\nu\in\left(-\infty,1\right)}.$$


As you correctly stated above, it can be shown that the distance between the two points in cylindrical coordinates is given by

$$\|\vec{r}_{0}-\vec{r}\|=\sqrt{\left(z_{0}-z\right)^{2}+\rho_{0}^{2}+\rho^{2}-2\rho_{0}\rho\cos{\left(\phi_{0}-\phi\right)}}.$$

The expression you gave for the vector quantity $\frac{\partial\vec{r}}{\partial\phi}\times\left(\vec{r}_{0}-\vec{r}\right)$, however, is wrong, so let's go over its calculation in detail.

Recall the cylindrical unit vectors $\hat{\rho}$ and $\hat{\phi}$ may be given in terms of Cartesian unit vectors by

$$\hat{\rho}=\cos{\left(\phi\right)}\,\hat{x}+\sin{\left(\phi\right)}\,\hat{y},$$

$$\hat{\phi}=-\sin{\left(\phi\right)}\,\hat{x}+\cos{\left(\phi\right)}\,\hat{y}.$$

Then, $\vec{r}=\rho\hat{\rho}+z\hat{z}$ and $\frac{\partial\vec{r}}{\partial\phi}=\rho\hat{\phi}$.

Similarly, $\vec{r}_{0}=\rho_{0}\hat{\rho}_{0}+z_{0}\hat{z}$, where the unit vectors $\hat{\rho}_{0}$ and $\hat{\phi}_{0}$ can be expressed as

$$\hat{\rho}_{0}=\cos{\left(\phi_{0}\right)}\,\hat{x}+\sin{\left(\phi_{0}\right)}\,\hat{y},$$

$$\hat{\phi}_{0}=-\sin{\left(\phi_{0}\right)}\,\hat{x}+\cos{\left(\phi_{0}\right)}\,\hat{y}.$$

The above pair of equations can be inverted to give $\hat{x}$ and $\hat{y}$ in terms of $\hat{\rho}_{0}$ and $\hat{\phi}_{0}$:

$$\hat{x}=\cos{\left(\phi_{0}\right)}\,\hat{\rho}_{0}-\sin{\left(\phi_{0}\right)}\,\hat{\phi}_{0},$$

$$\hat{y}=\sin{\left(\phi_{0}\right)}\,\hat{\rho}_{0}+\cos{\left(\phi_{0}\right)}\,\hat{\phi}_{0}.$$

We then find

$$\begin{align} \hat{\rho} &=\cos{\left(\phi\right)}\,\hat{x}+\sin{\left(\phi\right)}\,\hat{y}\\ &=\cos{\left(\phi\right)}\left[\cos{\left(\phi_{0}\right)}\,\hat{\rho}_{0}-\sin{\left(\phi_{0}\right)}\,\hat{\phi}_{0}\right]+\sin{\left(\phi\right)}\left[\sin{\left(\phi_{0}\right)}\,\hat{\rho}_{0}+\cos{\left(\phi_{0}\right)}\,\hat{\phi}_{0}\right]\\ &=\cos{\left(\phi_{0}-\phi\right)}\,\hat{\rho}_{0}-\sin{\left(\phi_{0}-\phi\right)}\,\hat{\phi}_{0},\\ \end{align}$$

$$\begin{align} \hat{\phi} &=-\sin{\left(\phi\right)}\,\hat{x}+\cos{\left(\phi\right)}\,\hat{y}\\ &=-\sin{\left(\phi\right)}\left[\cos{\left(\phi_{0}\right)}\,\hat{\rho}_{0}-\sin{\left(\phi_{0}\right)}\,\hat{\phi}_{0}\right]+\cos{\left(\phi\right)}\left[\sin{\left(\phi_{0}\right)}\,\hat{\rho}_{0}+\cos{\left(\phi_{0}\right)}\,\hat{\phi}_{0}\right]\\ &=\sin{\left(\phi_{0}-\phi\right)}\,\hat{\rho}_{0}+\cos{\left(\phi_{0}-\phi\right)}\,\hat{\phi}_{0}.\\ \end{align}$$

and thus,

$$\begin{align} \hat{\phi}\times\left(\vec{r}_{0}-\vec{r}\right) &=\hat{\phi}\times\left(\rho_{0}\hat{\rho}_{0}+z_{0}\hat{z}-\rho\hat{\rho}-z\hat{z}\right)\\ &=\hat{\phi}\times\left[\left(z_{0}-z\right)\hat{z}-\rho\hat{\rho}+\rho_{0}\hat{\rho}_{0}\right]\\ &=\left(z_{0}-z\right)\hat{\phi}\times\hat{z}-\rho\hat{\phi}\times\hat{\rho}+\rho_{0}\hat{\phi}\times\hat{\rho}_{0}\\ &=\left(z_{0}-z\right)\hat{\rho}+\rho\hat{z}-\rho_{0}\hat{\rho}_{0}\times\hat{\phi}\\ &=\left(z_{0}-z\right)\left[\cos{\left(\phi_{0}-\phi\right)}\,\hat{\rho}_{0}-\sin{\left(\phi_{0}-\phi\right)}\,\hat{\phi}_{0}\right]+\rho\hat{z}\\ &~~~~~-\rho_{0}\hat{\rho}_{0}\times\left[\sin{\left(\phi_{0}-\phi\right)}\,\hat{\rho}_{0}+\cos{\left(\phi_{0}-\phi\right)}\,\hat{\phi}_{0}\right]\\ &=\left(z_{0}-z\right)\cos{\left(\phi_{0}-\phi\right)}\,\hat{\rho}_{0}-\left(z_{0}-z\right)\sin{\left(\phi_{0}-\phi\right)}\,\hat{\phi}_{0}+\rho\hat{z}\\ &~~~~~-\rho_{0}\sin{\left(\phi_{0}-\phi\right)}\,\hat{\rho}_{0}\times\hat{\rho}_{0}-\rho_{0}\cos{\left(\phi_{0}-\phi\right)}\,\hat{\rho}_{0}\times\hat{\phi}_{0}\\ &=\left(z_{0}-z\right)\cos{\left(\phi_{0}-\phi\right)}\,\hat{\rho}_{0}-\left(z_{0}-z\right)\sin{\left(\phi_{0}-\phi\right)}\,\hat{\phi}_{0}+\rho\hat{z}\\ &~~~~~-\rho_{0}\cos{\left(\phi_{0}-\phi\right)}\,\hat{z}\\ &=\left(z_{0}-z\right)\cos{\left(\phi_{0}-\phi\right)}\,\hat{\rho}_{0}-\left(z_{0}-z\right)\sin{\left(\phi_{0}-\phi\right)}\,\hat{\phi}_{0}\\ &~~~~~+\left[\rho-\rho_{0}\cos{\left(\phi_{0}-\phi\right)}\right]\hat{z}.\\ \end{align}$$


We turn now to the calculation of $\vec{B}{\left(\vec{r}_{0}\right)}$ for the solenoid, and we start by simplifying the angular integral.

Given $\vec{r}_{0}\in\mathbb{R}^{3}$ such that $\neg\left[\rho_{0}=\rho\land\left(|z_{0}|\le\frac{L}{2}\right)\right]$, we have

$$\begin{align} \vec{B}{\left(\vec{r}_{0}\right)} &=n\int_{-\frac{L}{2}}^{\frac{L}{2}}\mathrm{d}z\,\frac{\mu_{0}I}{4\pi}\int_{0}^{2\pi}\mathrm{d}\phi\,\frac{\frac{\partial\vec{r}}{\partial\phi}\times\left(\vec{r}_{0}-\vec{r}\right)}{\|\vec{r}_{0}-\vec{r}\|^{3}}\\ &=\frac{\mu_{0}nI}{4\pi}\int_{-\frac{L}{2}}^{\frac{L}{2}}\mathrm{d}z\int_{0}^{2\pi}\mathrm{d}\phi\,\frac{\rho\hat{\phi}\times\left(\vec{r}_{0}-\vec{r}\right)}{\|\vec{r}_{0}-\vec{r}\|^{3}}\\ &=\frac{\mu_{0}nI}{4\pi}\int_{-\frac{L}{2}}^{\frac{L}{2}}\mathrm{d}z\int_{0}^{2\pi}\mathrm{d}\phi\,\frac{\rho}{\left[\left(z_{0}-z\right)^{2}+\rho_{0}^{2}+\rho^{2}-2\rho_{0}\rho\cos{\left(\phi_{0}-\phi\right)}\right]^{3/2}}\\ &~~~~~\times\bigg{[}\left(z_{0}-z\right)\cos{\left(\phi_{0}-\phi\right)}\,\hat{\rho}_{0}-\left(z_{0}-z\right)\sin{\left(\phi_{0}-\phi\right)}\,\hat{\phi}_{0}+\left[\rho-\rho_{0}\cos{\left(\phi_{0}-\phi\right)}\right]\hat{z}\bigg{]}\\ &=\frac{\mu_{0}nI}{4\pi}\int_{-\frac{L}{2}}^{\frac{L}{2}}\mathrm{d}z\int_{0}^{2\pi}\mathrm{d}\phi\,\frac{\rho}{\left[\left(z_{0}-z\right)^{2}+\rho_{0}^{2}+\rho^{2}-2\rho_{0}\rho\cos{\left(\phi_{0}-\phi\right)}\right]^{3/2}}\\ &~~~~~\times\bigg{[}\left(z_{0}-z\right)\cos{\left(\phi_{0}-\phi\right)}\,\hat{\rho}_{0}+\left[\rho-\rho_{0}\cos{\left(\phi_{0}-\phi\right)}\right]\hat{z}\bigg{]}\\ &=\frac{\mu_{0}nI}{4\pi}\int_{-\frac{L}{2}}^{\frac{L}{2}}\mathrm{d}z\int_{0}^{2\pi}\mathrm{d}\phi\,\frac{\rho}{\left[\left(z_{0}-z\right)^{2}+\rho_{0}^{2}+\rho^{2}-2\rho_{0}\rho\cos{\left(\phi\right)}\right]^{3/2}}\\ &~~~~~\times\bigg{[}\left(z_{0}-z\right)\cos{\left(\phi\right)}\,\hat{\rho}_{0}+\left[\rho-\rho_{0}\cos{\left(\phi\right)}\right]\hat{z}\bigg{]},\\ \end{align}$$

where in the second-to-last line above we used the fact that the $\hat{\phi}_{0}$ component of the angular integral is identically zero because of its $2\pi$-periodic antiderivative, and in the last line above we've used the following lemma: for any $a\in\mathbb{R}$ and any continuous periodic function $f:\mathbb{R}\rightarrow\mathbb{R}$ with period $p\in\mathbb{R}_{>0}$, it can be shown that

$$\int_{0}^{p}\mathrm{d}x\,f{\left(x+a\right)}=\int_{0}^{p}\mathrm{d}x\,f{\left(x\right)}.$$

Then,

$$\begin{align} \vec{B}{\left(\vec{r}_{0}\right)} &=\frac{\mu_{0}nI}{4\pi}\int_{-\frac{L}{2}}^{\frac{L}{2}}\mathrm{d}z\int_{0}^{2\pi}\mathrm{d}\phi\,\frac{\rho}{\left[\left(z_{0}-z\right)^{2}+\rho_{0}^{2}+\rho^{2}-2\rho_{0}\rho\cos{\left(\phi\right)}\right]^{3/2}}\\ &~~~~~\times\bigg{[}\left(z_{0}-z\right)\cos{\left(\phi\right)}\,\hat{\rho}_{0}+\left[\rho-\rho_{0}\cos{\left(\phi\right)}\right]\hat{z}\bigg{]}\\ &=\frac{\mu_{0}nI}{4\pi}\int_{z_{0}-\frac{L}{2}}^{z_{0}+\frac{L}{2}}\mathrm{d}z\int_{0}^{2\pi}\mathrm{d}\phi\,\frac{\rho}{\left[z^{2}+\rho_{0}^{2}+\rho^{2}-2\rho_{0}\rho\cos{\left(\phi\right)}\right]^{3/2}}\\ &~~~~~\times\bigg{[}z\cos{\left(\phi\right)}\,\hat{\rho}_{0}+\left[\rho-\rho_{0}\cos{\left(\phi\right)}\right]\hat{z}\bigg{]};~~~\small{\left[z\mapsto z_{0}-z\right]}\\ &=\frac{\mu_{0}nI}{4\pi}\int_{z_{0}-\frac{L}{2}}^{z_{0}+\frac{L}{2}}\mathrm{d}z\int_{0}^{\pi}\mathrm{d}\phi\,\frac{2\rho}{\left[z^{2}+\rho_{0}^{2}+\rho^{2}-2\rho_{0}\rho\cos{\left(\phi\right)}\right]^{3/2}}\\ &~~~~~\times\bigg{[}z\cos{\left(\phi\right)}\,\hat{\rho}_{0}+\left[\rho-\rho_{0}\cos{\left(\phi\right)}\right]\hat{z}\bigg{]};~~~\small{symmetry}\\ &=B_{0}\int_{z_{-}}^{z_{+}}\mathrm{d}z\int_{0}^{\pi}\mathrm{d}\phi\,\frac{2\rho\bigg{[}z\cos{\left(\phi\right)}\,\hat{\rho}_{0}+\left[\rho-\rho_{0}\cos{\left(\phi\right)}\right]\hat{z}\bigg{]}}{\left[z^{2}+\rho_{0}^{2}+\rho^{2}-2\rho_{0}\rho\cos{\left(\phi\right)}\right]^{3/2}},\\ \end{align}$$

where in the last line above we've set $B_{0}:=\frac{\mu_{0}nI}{4\pi}$ and $z_{\pm}:=z_{0}\pm\frac{L}{2}$.


Now, we could proceed to evaluate the angular integral in terms of elliptic integrals as others have already pointed out, but that would leave us with a very difficult non-elementary axial integral. It turns out, however, that changing the order of integration makes the integral much easier.

Consider the following derivative:

$$\frac{d}{dz}\left[\frac{z\vec{c}-a\vec{b}}{a\sqrt{z^{2}+a}}\right]=\frac{z\vec{b}+\vec{c}}{\left(z^{2}+a\right)^{3/2}};~~~\small{a\in\mathbb{R}_{>0}\land\vec{b}\in\mathbb{R}^{3}\land\vec{c}\in\mathbb{R}^{3}}.$$


Assuming $0<\rho_{0}\neq\rho$, we have $0<\rho_{0}^{2}+\rho^{2}-2\rho_{0}\rho\cos{\left(\phi\right)}$. Also note that $0<\frac{4\rho_{0}\rho}{z^{2}+\left(\rho_{0}+\rho\right)^{2}}<1$ for all real $z$.

Setting $\nu:=\frac{4\rho_{0}\rho}{\left(\rho_{0}+\rho\right)^{2}}\land k_{\pm}:=\sqrt{\frac{4\rho_{0}\rho}{z_{\pm}^{2}+\left(\rho_{0}+\rho\right)^{2}}}$, we then have

$$\begin{align} \vec{B}{\left(\vec{r}_{0}\right)} &=B_{0}\int_{z_{-}}^{z_{+}}\mathrm{d}z\int_{0}^{\pi}\mathrm{d}\phi\,\frac{2\rho\bigg{[}z\cos{\left(\phi\right)}\,\hat{\rho}_{0}+\left[\rho-\rho_{0}\cos{\left(\phi\right)}\right]\hat{z}\bigg{]}}{\left[z^{2}+\rho_{0}^{2}+\rho^{2}-2\rho_{0}\rho\cos{\left(\phi\right)}\right]^{3/2}}\\ &=B_{0}\int_{0}^{\pi}\mathrm{d}\phi\int_{z_{-}}^{z_{+}}\mathrm{d}z\,\frac{2\rho\bigg{[}z\cos{\left(\phi\right)}\,\hat{\rho}_{0}+\left[\rho-\rho_{0}\cos{\left(\phi\right)}\right]\hat{z}\bigg{]}}{\left[z^{2}+\rho_{0}^{2}+\rho^{2}-2\rho_{0}\rho\cos{\left(\phi\right)}\right]^{3/2}}\\ &=B_{0}\int_{0}^{\pi}\mathrm{d}\phi\int_{z_{-}}^{z_{+}}\mathrm{d}z\,\frac{d}{dz}\bigg{[}-\frac{2\rho\cos{\left(\phi\right)}\,\hat{\rho}_{0}}{\sqrt{z^{2}+\rho_{0}^{2}+\rho^{2}-2\rho_{0}\rho\cos{\left(\phi\right)}}}\\ &~~~~~+\frac{2z\rho\left[\rho-\rho_{0}\cos{\left(\phi\right)}\right]\hat{z}}{\left[\rho_{0}^{2}+\rho^{2}-2\rho_{0}\rho\cos{\left(\phi\right)}\right]\sqrt{z^{2}+\rho_{0}^{2}+\rho^{2}-2\rho_{0}\rho\cos{\left(\phi\right)}}}\bigg{]}\\ &=B_{0}\int_{0}^{\pi}\mathrm{d}\phi\,\bigg{[}-\frac{2\rho\cos{\left(\phi\right)}\,\hat{\rho}_{0}}{\sqrt{z_{+}^{2}+\rho_{0}^{2}+\rho^{2}-2\rho_{0}\rho\cos{\left(\phi\right)}}}+\frac{2\rho\cos{\left(\phi\right)}\,\hat{\rho}_{0}}{\sqrt{z_{-}^{2}+\rho_{0}^{2}+\rho^{2}-2\rho_{0}\rho\cos{\left(\phi\right)}}}\\ &~~~~~+\frac{2z_{+}\rho\left[\rho-\rho_{0}\cos{\left(\phi\right)}\right]\hat{z}}{\left[\rho_{0}^{2}+\rho^{2}-2\rho_{0}\rho\cos{\left(\phi\right)}\right]\sqrt{z_{+}^{2}+\rho_{0}^{2}+\rho^{2}-2\rho_{0}\rho\cos{\left(\phi\right)}}}\\ &~~~~~-\frac{2z_{-}\rho\left[\rho-\rho_{0}\cos{\left(\phi\right)}\right]\hat{z}}{\left[\rho_{0}^{2}+\rho^{2}-2\rho_{0}\rho\cos{\left(\phi\right)}\right]\sqrt{z_{-}^{2}+\rho_{0}^{2}+\rho^{2}-2\rho_{0}\rho\cos{\left(\phi\right)}}}\bigg{]}\\ &=2B_{0}\int_{0}^{\frac{\pi}{2}}\mathrm{d}\varphi\,\bigg{[}-\frac{2\rho\cos{\left(2\varphi\right)}\,\hat{\rho}_{0}}{\sqrt{z_{+}^{2}+\rho_{0}^{2}+\rho^{2}-2\rho_{0}\rho\cos{\left(2\varphi\right)}}}\\ &~~~~~+\frac{2\rho\cos{\left(2\varphi\right)}\,\hat{\rho}_{0}}{\sqrt{z_{-}^{2}+\rho_{0}^{2}+\rho^{2}-2\rho_{0}\rho\cos{\left(2\varphi\right)}}}\\ &~~~~~+\frac{2z_{+}\rho\left[\rho-\rho_{0}\cos{\left(2\varphi\right)}\right]\hat{z}}{\left[\rho_{0}^{2}+\rho^{2}-2\rho_{0}\rho\cos{\left(2\varphi\right)}\right]\sqrt{z_{+}^{2}+\rho_{0}^{2}+\rho^{2}-2\rho_{0}\rho\cos{\left(2\varphi\right)}}}\\ &~~~~~-\frac{2z_{-}\rho\left[\rho-\rho_{0}\cos{\left(2\varphi\right)}\right]\hat{z}}{\left[\rho_{0}^{2}+\rho^{2}-2\rho_{0}\rho\cos{\left(2\varphi\right)}\right]\sqrt{z_{-}^{2}+\rho_{0}^{2}+\rho^{2}-2\rho_{0}\rho\cos{\left(2\varphi\right)}}}\bigg{]};~~~\small{\left[\phi=2\varphi\right]}\\ &=2B_{0}\int_{0}^{\frac{\pi}{2}}\mathrm{d}\varphi\,\bigg{[}\frac{\left[2\rho-4\rho\cos^{2}{\left(\varphi\right)}\right]\hat{\rho}_{0}}{\sqrt{z_{+}^{2}+\left(\rho_{0}+\rho\right)^{2}-4\rho_{0}\rho\cos^{2}{\left(\varphi\right)}}}\\ &~~~~~-\frac{\left[2\rho-4\rho\cos^{2}{\left(\varphi\right)}\right]\hat{\rho}_{0}}{\sqrt{z_{-}^{2}+\left(\rho_{0}+\rho\right)^{2}-4\rho_{0}\rho\cos^{2}{\left(\varphi\right)}}}\\ &~~~~~+\frac{\left[-\rho_{0}^{2}+\rho^{2}+\left(\rho_{0}+\rho\right)^{2}-4\rho_{0}\rho\cos^{2}{\left(\varphi\right)}\right]z_{+}\hat{z}}{\left[\left(\rho_{0}+\rho\right)^{2}-4\rho_{0}\rho\cos^{2}{\left(\varphi\right)}\right]\sqrt{z_{+}^{2}+\left(\rho_{0}+\rho\right)^{2}-4\rho_{0}\rho\cos^{2}{\left(\varphi\right)}}}\\ &~~~~~-\frac{\left[-\rho_{0}^{2}+\rho^{2}+\left(\rho_{0}+\rho\right)^{2}-4\rho_{0}\rho\cos^{2}{\left(\varphi\right)}\right]z_{-}\hat{z}}{\left[\left(\rho_{0}+\rho\right)^{2}-4\rho_{0}\rho\cos^{2}{\left(\varphi\right)}\right]\sqrt{z_{-}^{2}+\left(\rho_{0}+\rho\right)^{2}-4\rho_{0}\rho\cos^{2}{\left(\varphi\right)}}}\bigg{]}\\ &=\frac{B_{0}}{\sqrt{\rho_{0}\rho}}\int_{0}^{\frac{\pi}{2}}\mathrm{d}\varphi\,\bigg{[}\frac{2\rho\left[1-2\cos^{2}{\left(\varphi\right)}\right]k_{+}\hat{\rho}_{0}}{\sqrt{1-k_{+}^{2}\cos^{2}{\left(\varphi\right)}}}-\frac{2\rho\left[1-2\cos^{2}{\left(\varphi\right)}\right]k_{-}\hat{\rho}_{0}}{\sqrt{1-k_{-}^{2}\cos^{2}{\left(\varphi\right)}}}\\ &~~~~~+\frac{\left[-\left(\frac{\rho_{0}-\rho}{\rho_{0}+\rho}\right)+1-\nu\cos^{2}{\left(\varphi\right)}\right]k_{+}z_{+}\hat{z}}{\left[1-\nu\cos^{2}{\left(\varphi\right)}\right]\sqrt{1-k_{+}^{2}\cos^{2}{\left(\varphi\right)}}}\\ &~~~~~-\frac{\left[-\left(\frac{\rho_{0}-\rho}{\rho_{0}+\rho}\right)+1-\nu\cos^{2}{\left(\varphi\right)}\right]k_{-}z_{-}\hat{z}}{\left[1-\nu\cos^{2}{\left(\varphi\right)}\right]\sqrt{1-k_{-}^{2}\cos^{2}{\left(\varphi\right)}}}\bigg{]},\\ \end{align}$$

and finally,

$$\begin{align} \vec{B}{\left(\vec{r}_{0}\right)} &=\frac{B_{0}}{\sqrt{\rho_{0}\rho}}\int_{0}^{\frac{\pi}{2}}\mathrm{d}\varphi\,\bigg{[}\frac{2\rho\left[1-2\sin^{2}{\left(\varphi\right)}\right]k_{+}\hat{\rho}_{0}}{\sqrt{1-k_{+}^{2}\sin^{2}{\left(\varphi\right)}}}-\frac{2\rho\left[1-2\sin^{2}{\left(\varphi\right)}\right]k_{-}\hat{\rho}_{0}}{\sqrt{1-k_{-}^{2}\sin^{2}{\left(\varphi\right)}}}\\ &~~~~~+\frac{\left[-\left(\frac{\rho_{0}-\rho}{\rho_{0}+\rho}\right)+1-\nu\sin^{2}{\left(\varphi\right)}\right]k_{+}z_{+}\hat{z}}{\left[1-\nu\sin^{2}{\left(\varphi\right)}\right]\sqrt{1-k_{+}^{2}\sin^{2}{\left(\varphi\right)}}}\\ &~~~~~-\frac{\left[-\left(\frac{\rho_{0}-\rho}{\rho_{0}+\rho}\right)+1-\nu\sin^{2}{\left(\varphi\right)}\right]k_{-}z_{-}\hat{z}}{\left[1-\nu\sin^{2}{\left(\varphi\right)}\right]\sqrt{1-k_{-}^{2}\sin^{2}{\left(\varphi\right)}}}\bigg{]};~~~\small{\left[\varphi\mapsto\frac{\pi}{2}-\varphi\right]}\\ &=\frac{B_{0}}{\sqrt{\rho_{0}\rho}}\int_{0}^{\frac{\pi}{2}}\mathrm{d}\varphi\,\frac{2\rho\left[1-2\sin^{2}{\left(\varphi\right)}\right]k_{+}\hat{\rho}_{0}}{\sqrt{1-k_{+}^{2}\sin^{2}{\left(\varphi\right)}}}\\ &~~~~~-\frac{B_{0}}{\sqrt{\rho_{0}\rho}}\int_{0}^{\frac{\pi}{2}}\mathrm{d}\varphi\,\frac{2\rho\left[1-2\sin^{2}{\left(\varphi\right)}\right]k_{-}\hat{\rho}_{0}}{\sqrt{1-k_{-}^{2}\sin^{2}{\left(\varphi\right)}}}\\ &~~~~~+\frac{B_{0}}{\sqrt{\rho_{0}\rho}}\int_{0}^{\frac{\pi}{2}}\mathrm{d}\varphi\,\frac{\left[-\left(\frac{\rho_{0}-\rho}{\rho_{0}+\rho}\right)+1-\nu\sin^{2}{\left(\varphi\right)}\right]k_{+}z_{+}\hat{z}}{\left[1-\nu\sin^{2}{\left(\varphi\right)}\right]\sqrt{1-k_{+}^{2}\sin^{2}{\left(\varphi\right)}}}\\ &~~~~~-\frac{B_{0}}{\sqrt{\rho_{0}\rho}}\int_{0}^{\frac{\pi}{2}}\mathrm{d}\varphi\,\frac{\left[-\left(\frac{\rho_{0}-\rho}{\rho_{0}+\rho}\right)+1-\nu\sin^{2}{\left(\varphi\right)}\right]k_{-}z_{-}\hat{z}}{\left[1-\nu\sin^{2}{\left(\varphi\right)}\right]\sqrt{1-k_{-}^{2}\sin^{2}{\left(\varphi\right)}}}\\ &=\hat{\rho}_{0}\frac{4B_{0}\sqrt{\rho}}{k_{+}\sqrt{\rho_{0}}}\int_{0}^{\frac{\pi}{2}}\mathrm{d}\varphi\,\frac{\left[1-k_{+}^{2}\sin^{2}{\left(\varphi\right)}\right]-\left(1-\frac{k_{+}^{2}}{2}\right)}{\sqrt{1-k_{+}^{2}\sin^{2}{\left(\varphi\right)}}}\\ &~~~~~-\hat{\rho}_{0}\frac{4B_{0}\sqrt{\rho}}{k_{-}\sqrt{\rho_{0}}}\int_{0}^{\frac{\pi}{2}}\mathrm{d}\varphi\,\frac{\left[1-k_{-}^{2}\sin^{2}{\left(\varphi\right)}\right]-\left(1-\frac{k_{-}^{2}}{2}\right)}{\sqrt{1-k_{-}^{2}\sin^{2}{\left(\varphi\right)}}}\\ &~~~~~+\hat{z}\frac{B_{0}k_{+}z_{+}}{\sqrt{\rho_{0}\rho}}\int_{0}^{\frac{\pi}{2}}\mathrm{d}\varphi\,\frac{\left[1-\nu\sin^{2}{\left(\varphi\right)}\right]-\left(\frac{\rho_{0}-\rho}{\rho_{0}+\rho}\right)}{\left[1-\nu\sin^{2}{\left(\varphi\right)}\right]\sqrt{1-k_{+}^{2}\sin^{2}{\left(\varphi\right)}}}\\ &~~~~~-\hat{z}\frac{B_{0}k_{-}z_{-}}{\sqrt{\rho_{0}\rho}}\int_{0}^{\frac{\pi}{2}}\mathrm{d}\varphi\,\frac{\left[1-\nu\sin^{2}{\left(\varphi\right)}\right]-\left(\frac{\rho_{0}-\rho}{\rho_{0}+\rho}\right)}{\left[1-\nu\sin^{2}{\left(\varphi\right)}\right]\sqrt{1-k_{-}^{2}\sin^{2}{\left(\varphi\right)}}}\\ &=\frac{4B_{0}\sqrt{\rho}}{k_{+}\sqrt{\rho_{0}}}\bigg{[}E{\left(k_{+}\right)}-\left(1-\frac{k_{+}^{2}}{2}\right)K{\left(k_{+}\right)}\bigg{]}\hat{\rho}_{0}\\ &~~~~~-\frac{4B_{0}\sqrt{\rho}}{k_{-}\sqrt{\rho_{0}}}\bigg{[}E{\left(k_{-}\right)}-\left(1-\frac{k_{-}^{2}}{2}\right)K{\left(k_{-}\right)}\bigg{]}\hat{\rho}_{0}\\ &~~~~~+\frac{B_{0}k_{+}z_{+}}{\sqrt{\rho_{0}\rho}}\bigg{[}K{\left(k_{+}\right)}-\left(\frac{\rho_{0}-\rho}{\rho_{0}+\rho}\right)\Pi{\left(\nu,k_{+}\right)}\bigg{]}\hat{z}\\ &~~~~~-\frac{B_{0}k_{-}z_{-}}{\sqrt{\rho_{0}\rho}}\bigg{[}K{\left(k_{-}\right)}-\left(\frac{\rho_{0}-\rho}{\rho_{0}+\rho}\right)\Pi{\left(\nu,k_{-}\right)}\bigg{]}\hat{z}.\blacksquare\\ \end{align}$$


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  • $\begingroup$ Thank you for sharing the answer. It is a huge help. Will take some time, to read and understand it. $\endgroup$ Commented Jun 7, 2022 at 18:58
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    $\begingroup$ I appreciate the effort, to write it in such detail. $\endgroup$ Commented Jun 7, 2022 at 19:00
  • $\begingroup$ Finally I get it. It was a very decent method. Thank you $\endgroup$ Commented Jul 3, 2022 at 14:36
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Let $t=x-x_0$, $c=\frac ab$ and write $$\int\frac{m+n \cos (t)}{\big[a+b \cos (t)\big]^{3/2}}\,dt=\frac n {b^{3/2}} \int\frac{dt}{\big[c+ \cos (t)\big]^{1/2}}+\frac{ mb-an}{b^{5/2} }\int\frac{dt}{\big[c+ \cos (t)\big]^{3/2}}$$ $$I_1=\int\frac{dt}{\big[c+ \cos (t)\big]^{1/2}}=\frac{2 }{\sqrt{c+1}}\,\,F\left(\frac{t}{2}|\frac{2}{c+1}\right)$$ $$I_2=\int\frac{dt}{\big[c+ \cos (t)\big]^{3/2}}=\frac{2 \sqrt{c+1}}{c^2-1}\,\, E\left(\frac{t}{2}|\frac{2}{c+1}\right)-\frac{2 }{\left(c^2-1\right) }\frac{ \sin (t)}{\sqrt{c+\cos (t)}}$$ where appear the elliptic integrals of the first and second kinds.

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  • $\begingroup$ I know this result, and I have calculated my definite integral by this method. Here we are hence left with only, with ellptical integral of form $\mathrm{E}(\pi,k)$ and $\mathrm{F}(\pi,k)$. $\endgroup$ Commented Mar 30, 2022 at 13:12
  • $\begingroup$ I was asking for the second expression, i.e integration of $\vec{B}$ that involves integration of elliptical function. $\endgroup$ Commented Mar 30, 2022 at 13:15
  • $\begingroup$ @NikolaAlfredi. If $t$ is small, why not to use series expansions for $I_1$ and $I_2$ $\endgroup$ Commented Mar 30, 2022 at 13:21
  • $\begingroup$ $t$ is not small cause it vary from $0$ to $2\pi.$ $\endgroup$ Commented Mar 30, 2022 at 13:26
  • $\begingroup$ Am asking $\displaystyle \int_{z_i}^{z^f} \frac {(z_0-z) ( ( z-z_0)^2 +\rho^2 + \rho_0^2 ) }{\sqrt{( z-z_0)^2 + (\rho - \rho_0)^2} (( z-z_0)^2 + (\rho + \rho_0)^2) } \mathrm{E} \left( \pi, \frac{4\rho \rho_0}{ (z-z_0)^2 + (\rho - \rho_0)^2 } \right) \mathrm{d}z $ $\endgroup$ Commented Mar 30, 2022 at 13:32

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