# For closed circuits, why can't we have more than one $f(r)$?

Force between current elements depends on a function of angles [$f(\eta, \theta, \theta^{\prime})$] and also on a function of distance between them [$f(r)$] .

For closed circuits, there are more than one valid $f(\eta, \theta, \theta^{\prime})$ like: \begin{align} \vec{F} & =k \ ii'\oint^{s^\prime}_0 \oint^s_0\dfrac{\vec{ds}\times (\vec{ds^{\prime}} \times \hat{r})}{r^2}\\ & =-k \ ii'\oint^{s^\prime}_0 \oint^s_0 \dfrac{2(\vec{ds}.\vec{ds^\prime})-3(\hat{r}.\vec{ds})(\hat{r}.\vec{ds^{\prime}})}{r^2}(\hat{r})\\ & =-k \ ii'\oint^{s^\prime}_0 \oint^s_0 \dfrac{\vec{ds}. \vec{ds^{\prime}}}{r^2} (\hat{r}) \end{align}

However for closed circuits, there is only one valid $f(r)$, i.e. $\dfrac{1}{r^2}$.

Why is it impossible for closed circuits to have more than one $f(r)$ while they can have more than one $f(\eta, \theta, \theta^{\prime})$?

• Hello!!! Is anyone here??? – N.G.Tyson Apr 23 '18 at 11:17
• You can't expect instantaneous answers, the content on this site is provided by volunteers who supply it in their spare time, don't expect an answer within two hours. The comment does not really increase your chances of getting an answer. Also, your question is not very clearly stated using $f$ for two different entities is not good practice. – Sebastian Riese Apr 23 '18 at 15:46

Because we can always rewrite formulas to yield equivalent expressions. For closed line integrals we can, for example, add an arbitrary gradient $\nabla \phi$ to the integrand without changing the result. We can also use algebraic identities to reformulate the integrands (for example the identitiy $\vec a \times ( \vec b \times \vec c) = \vec b (\vec a \cdot \vec c) - \vec c (\vec a \cdot \vec b)$).
We could do the same for the $1/r^2$ (e.g. we could use $1/r^3$ and use $\vec r$ instead of $\hat r = \vec r/r$ in the numerator).