Question

$N={\mathrm{2.5.10}}^3$ 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 $\eta$ of the magnetic induction inside the core to that at the centre of the toroid.

Answer 1

Magnetic field inside toroid$=\frac{{\mu }_{0}NI}{2\pi r}$
The magnetic field at center is non zero as each loop is not a closed circle but it is a helix
For each turn we move by $\frac{2\pi r}{N}$ in toroidal diretion
There are two components of current :one along x and one along y
Magnetic field due to $I_2$ (along circle) at center is zero
But due to $I_1$ there is a magnetic field at center
So by Biot-Savarts Law
$B_{center}=\frac{{\mu }_0}{4\pi }\int \frac{Idl\times \hat{r}}{r^2}$
$=\frac{{\mu }_0}{4\pi }I\times \frac{2\pi r}{r^2}$
$B_{center}=\frac{{\mu }_0}{4\pi }I\frac{2\pi r}{r}=\frac{{\mu }_{0}I}{2r}$
So $\eta =\frac{B_{inside}}{B_{center}}=\frac{\frac{{\mu }_{0}NI}{2\pi r}}{\frac{{\mu }_{0}I}{2r}}$
$=\frac{N}{\pi }=\frac{25}{\pi }\times {10}^2$
$\eta \approx 8\times {10}^2$
$\Rightarrow x=8$