According to Ampere’s law $$\oint \vec{B}. d\vec{l}=\mu_0I_e$$
Where, $I_e$ is the current enclosed by the amperian loop
So if I consider an amperian loop at the centre of the circular current carrying wire(with the same centre) with a radius lesser than that of the current carrying loop, since there is no current enclosed($I_e$) in the loop there must be no magnetic field at the centre.
But there is a formula for magnetic field due to current carrying arcs which gives us the expression
$$\vec{B}=\frac{\mu_0 I \theta}{4\pi r}$$
We can get the magnetic field by putting $\theta=2\pi$
Why is my initial statement wrong?
Best Answer
Ampere's law:
$$\oint_C \vec{B} \cdot d\vec{l}=\mu_0I_e$$
The integral tells us that if the magnetic field is zero at every point along the closed curve C, then there must be no net current enclosed in the loop.
However, the converse is not true -- there are many ways for the magnetic field to be non-zero at most or every point along that curve, and yet still have the total integral evaluate to a total of zero.
In other words, there are many ways to have a strong magnetic field in free space, even when a loop in that region does not include any current.
With an Amperian loop at the centre of a circular current carrying wire with a radius lesser than that of the current carrying loop, yes there is no current enclosed(Ie) in the loop.
However, there is a strong magnetic field pointing through the current carrying loop of wire. When actually doing the integration, the little "dl" vectors are in the plane of the loop of wire, but the magnetic field "B" is at right angles to that plane, so the dot product is zero.
Let's now consider a slightly different Amperian loop, perhaps one still centered on the center of the current-carrying loop of wire and still smaller than that loop, but now tilted at some angle. Since the little "dl" vectors are not exactly at right angles to the "B" field generated by that current, the dot product is not zero, so there is a part of that curve where the integral is positive. However, the integral along the rest of the curve is negative, so the total integral still evaluates to a total of zero.