Question

A toroid having a square cross section, $\mathbf{5}\mathbf{.}\mathbf{00}\mathbf{}\mathbf{cm}$on a side, and an inner radius of$\mathbf{15}\mathbf{.}\mathbf{0}\mathbf{}\mathbf{cm}$has$\mathbf{500}\mathbf{}\mathbf{turns}$and carries a current of$\mathbf{0}\mathbf{.}\mathbf{800}\mathbf{}\mathbf{A}$. (It is made up of a square solenoid—instead of a round one as in Figure bent into a doughnut shape.) (a) What is the magnetic field inside the toroid at the inner radius and (b) What is the magnetic field inside the toroid at the outer radius?

Short answer

a) The magnetic field inside the toroid at the inner radiusis$5.33\times {10}^{-4}\mathrm{T}$

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

Step-by-step solution

Listing the given quantities

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

$\mathrm{N}=500$

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

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

Understanding the concept of magnetic field and toroid

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

$\mathbf{B}\mathbf{=}\frac{{\mathbf{\mu }}_{\mathbf{0}}\mathbf{iN}}{\mathbf{2}\mathbf{\pi }}\frac{\mathbf{1}}{\mathbf{r}}\mathbf{}$(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.

(a) Calculations of the magnetic field inside the toroid at the inner radius

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

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

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

(b) Calculations of the magnetic field inside the toroid at the outer radius

Similarly, for outer radius we have,

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

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