Magnetic Induction at the centre of a Toroid This problem is in reference to Q3.239 in Problems in General Physics by I.E. Irodov.
Question:  N = $2.5*10^3 $wire turns are uniformly wound on a wooden toroidal core of very small cross-section. A current I flows through the wire. Find the ratio 1 of the magnetic induction inside the core to that at the centre of the toroid. 
Assuming an Amperian loop in the central space of toroid directly gives us B = 0 at .the center, similarly inside the toroid gives us $$ B = \mu NI / (2 \pi r) $$ and this is also what I have found in many reference books too. But the answer given is $$ \eta = N / \pi$$
Where have i missed?
Thanks.
 A: You are correct in your calculation of magnetic field inside the toroid, it is indeed $$ B_\text{inside} = \mu NI / (2 \pi r).$$
However, in the center the magnetic field is non-zero. Indeed, each loop is not a closed circle but a turn of a helix. Each such turn contain a shift of $2\pi r /N$ in toroidal direction. And so while poloidal components of the current do no contribute to the magnetic field at the center, the toroidal components do. And their contributions (for thin toroid) equivalent to that of single circular coil in the toroidal direction. (For definition of directions on torus look here).
Thus, by Biot-Savart law, at the center magnetic field would be
$$
\mathbf{B}_\text{center} = \frac{\mu}{4 \pi} \int_C \frac{I d{\boldsymbol\ell} \times \hat{\mathbf{r}}}{r^2}=\hat{\mathbf{n}}\, \frac{\mu}{4\pi} \,\frac{ 2\pi r\, I }{r^2}=\hat{ \mathbf{n}}\,\frac{\mu I}{2 r},
$$
where we have taken into account that the direction of vector product is the same at each point on the circle.
And so the ratio is
$$ \eta =\frac{B_\text{inside}}{B_\text{center}}=\frac N \pi.$$
A: 
In above Figure-01 we have a circle of radius $\:R\:$ on the $\:xy-$plane with center on the origin $\:\rm O$. A circular loop of radius $\:r<R\:$ lies on a plane normal to the $\:xy-$plane with center on the circumference of the first circle. A steady current $\:I\:$ flows around the circular loop producing at the center a magnetic field $\:\mathbf{B}\:$ lying entirely on the $\:xy-$plane. The circular loop represents one of the $\:N\:$ turns of an ideal toroid (a wire wound around a torus). 


Now, if a number of identical circular loops are placed in a regular polygon arrangement then magnetic field at the center is canceled out. Figure-02 is an example of a regular pentagon arrangement.


The greater the number of turns $\:N\:$  the closer the ideal toroid to the real one. Figure-03 shows an ideal toroid with $\:N=72\:$ turns. The magnetic field  at the center is zero. 


Now, in Figure-04 we tilt the circular loop of Figure-01 by an angle $\:\theta\:$ around the $\:y-$axis because of the fact that the turns of a real toroid are inclined with respect to the $\:xy-$plane. In this case the magnetic field at the center has a component $\:\mathbf{B}_{\boldsymbol{\perp}}$ normal to the $\:xy-$plane.


In Figure-05 we see a regular polygon arrangement with  $\:N=72\:$ turns. The resulting magnetic field $\:\sum\mathbf{B}_{\boldsymbol{\perp}}$ at the center is  normal to the $\:xy-$plane.
