# Magnetic field due to a current-carrying loop at the wire

Consider a current-carrying loop. I am trying to work out the magnetic field at the a certain point on the wire due to the other points of the wire. I will call this point P. I am trying to calculate this to find out the effect of the current on the change in radius of the loop. It seems quite obvious that the radius will increase because the magnetic force will act outwards on the wire.

Using Biot-Savart law, I eventually reached the following equation:

$$B = \frac{\mu_0I}{4R\pi} \int_0^\pi \!\frac{1}{1- \cos\theta}\,d\theta$$

where the direction of B is obtained via right-hand grip rule, and $$\theta$$ is the angle subtended by the line from the centre to P and the line from the centre to another point along the wire that I am integrating over

The result is infinity.

Upon relooking at the problem, I realised that this could be because I am considering the effect of the magnetic field from parts of the wire very very close to point P.

And then, because of the inverse square relationship between $$|r|$$ and $$B$$ in the original Biot-Savart law, and $$|r|$$ tends to 0, then $$B$$ tends to infinity.

But I also see that such points should have a $$\theta$$ that tends to 0, resulting in a $$\sin\theta$$ (from $$dl \times \hat{r}$$) that tends to 0. I think this and the aforementioned inverse square effect should cancel out.

So why am I still getting infinity? How I do I resolve this problem/contradiction in the qualitative understanding of the problem? (I am not so concerned over the equation)

PS: Yes, I know there seem to be other questions asking the same thing, but those either do not consider a loop, or are focusing on a different effect (individual electrons in the wire having an effect on each other)

PPS: I apologise if my understanding of calculus may not be very good

Biot-Savart does produce such discontinuities unless one integrates in 3 dimensions. The full Biot-Savart equation is $$B(r)=\frac{\mu_0}{2\pi}\int\int\int_V \frac{(J \times \hat r)dV}{|r|^2}$$, a three dimensional integral. If one considers a infinitesimal cylindrical shell around a wire with some radius $$R$$, and some length $$dL$$ and some thickness $$dT$$, its volume is $$\pi R^2\ dL\ dT$$ and we can that the contribution of that shell to the final integral will approach $$\pi(J \times \hat r)\ dL$$, with no pesky $$|r|$$ in the denominator.
If you think about the 2d case, consider that what you're really computing is an infinitely thin ribbon of perfect conductor -- a construct which easily generates discontinuities. That nice, straight forward cylindrical shell gets stretched in one direction infinitely far, and we no longer get the $$R^2$$ term we need.