So I have this vector field
$$ \textbf{B}=K \left( \frac{\cos \varphi}{\rho^2}\textbf{e}_{\rho}+ \left( \frac{\sin \varphi}{\rho^2}+ \frac{1}{a\rho}\textbf{e}_{\varphi} \right) \right) $$ and the first part of the problem is to calculate a potential which I have done. So the potential is $$ \phi(\rho,\varphi,z)=-K\frac{\cos\varphi}{\rho}+\frac{K}{a}\varphi $$ Now to the problem. I'm asked to calculate the line integral
$$ \int_C \textbf{B} \cdot d\textbf{r} $$ where $C: \textbf{r}(s)=(b\cos(s),b\sin(s),b), \, 0≤s≤2\pi, \, b>0$
I can't do this with the potential, because of the singularities, right? When I try to compute the line integral by the parametrization it just gets messy and I can't integrate it. Can anyone show me how to do this last part or point me in the right direction?
I will write my interpretation of how I understand it. We have to calculate the work of the vector field (=the line integral) along the path $C$.
Method 1: Direct parametrization (by definition). $$ \textbf{B} \cdot d\textbf{r}=[\text{in cyl. coord.}]=(B_\rho, B_\phi,B_z)\cdot (d\rho,\rho\,d\varphi,dz)=K\Bigl(\frac{\cos\varphi}{\rho^2}\,d\rho+\Bigl(\frac{\sin\varphi}{\rho}+\frac{1}{a}\Bigr)\,d\varphi\Bigr). $$ Along the path we have $\rho=b$, that is $d\rho=0$, and $\varphi=s$, hence, the integral becomes $$ K\int_0^{2\pi}\Bigl(\frac{\sin s}{b}+\frac{1}{a}\Bigr)\,ds=K\int_0^{2\pi}\frac{1}{a}\,ds=\frac{2\pi K}{a}. $$ Method 2: Using the potential. Since in the new variables the vector field and the potential has no singularities along the path, we have that $$ \int_C \textbf{B} \cdot d\textbf{r}=\phi(\text{end point})-\phi(\text{start point})=\phi(b,2\pi,b)-\phi(b,0,b)=\frac{K}{a}2\pi. $$ Interpretation: The field can be split into two $$ \textbf{B}=\textbf{B}_1+\textbf{B}_2 $$ where $$ \textbf{B}_1=K \left( \frac{\cos \varphi}{\rho^2}\textbf{e}_{\rho}+\frac{\sin \varphi}{\rho^2}\textbf{e}_{\varphi}\right)=[xyz-\text{coord}]= \left(\frac{x^2-y^2}{(x^2+y^2)^2},\frac{2xy}{(x^2+y^2)^2},0\right) $$ which is a potential field with the potential $-K\frac{\cos\varphi}{\rho}=-K\frac{x}{x^2+y^2}$ (I am not very sure, but it looks like some kind of vector field of the electrostatic line dipole), and $$ \textbf{B}_2=K\frac{1}{a\rho}\textbf{e}_{\varphi}=[xyz-\text{coord}]=\left(\frac{-y}{x^2+y^2},\frac{x}{x^2+y^2},0\right) $$ which is the magnetic field around the infinite wire along the $z$ axis. The first field has zero work since the potential does not depend on $z$ and the start and the end points of the curve have the same $x,y$ coordinates. So the only contribution to the work comes from the second field with the known property that the work is proportional to the number of rotations around the wire times $2\pi$.