91Ó°ÊÓ

${\mathrm{r}}_{\mathrm{in}}=15\mathrm{cm}=0.15\mathrm{m}$

$\mathrm{N}=500$

$\begin{array}{c}{\mathrm{r}}_{\mathrm{out}}=15\mathrm{cm}+5\mathrm{cm}\\ =20\mathrm{cm}\\ =0.20\mathrm{m}\end{array}$

$\mathrm{i}=0.800\mathrm{A}$

At a point inside the toroid, the magnitude $\mathbf{B}$of the magnetic field is given by equation,

$\mathbf{B}\mathbf{=}\frac{{\mathbf{μ}}_{\mathbf{0}}\mathbf{iN}}{\mathbf{2}\mathbf{π}}\frac{\mathbf{1}}{\mathbf{r}}\mathbf{}$(i)

Nis the number of turns and r is the distance of the point from the center of the toroid.

Using that equation, we can calculate the magnetic field inside the toroid at the inner and outer radius.

By using equation (i), we can calculate the magnetic field at the inner radius,

$\begin{array}{c}\mathrm{B}=\frac{{\mathrm{μ}}_0\mathrm{iN}}{2\mathrm{π}}\frac{1}{{\mathrm{r}}_{\mathrm{in}}}\\ =\frac{1.26×{10}^{−6}×0.800×500}{6.28}\frac{1}{0.15}\\ =5.33×{10}^{-4}\mathrm{T}\end{array}$

Thus, the magnetic field inside the toroid at the inner radius is $5.33×{10}^{-4}\mathrm{T}$.

Similarly, for outer radius we have,

$\begin{array}{c}\mathrm{B}=\frac{{\mathrm{μ}}_0\mathrm{iN}}{2\mathrm{π}}\frac{1}{{\mathrm{r}}_{\mathrm{out}}}\\ =\frac{1.26×{10}^{−6}×0.800×500}{6.28}\frac{1}{0.20}\\ =4.00×{10}^{-4}\mathrm{T}\end{array}$

The magnetic field inside the toroid at the outer radius is$4.00×{10}^{-4}\mathrm{T}$