A toroid has a 5.00 cmsquare cross section, an inside radius of 0.15m, 500turns of wire, and a current of 0.800A. What is the magnetic flux through the cross section?

$$\begin{gathered} \mathrm{Inner}\mathrm{radius}=15\mathrm{cm}=0.15\mathrm{m} \\ N=500 \\ b=5.00\mathrm{cm}=0.05\mathrm{m} \end{gathered}$$

By using the expression$i=0.800\mathrm{A}$for magnetic field of toroid in the formula for flux and integrating it, we can find the total magnetic flux through the cross-section of the toroid.

Formula:

$$\begin{gathered} B=\frac{{\mu }_{0}ni}{2\mathrm{\pi r}} \\ d\varphi =BdA \end{gathered}$$

Suppose the toroid has a square cross-section with side b and inner radius. If current i flows in the toroid the magnetic field at a distance r , such that $a<r<a+b$is given by

$B=\frac{{\mu }_{0}Ni}{2\pi r}$

Assuming that the magnetic field is non-uniform over the cross section of the toroid, we take an area element of width dr and height b. The flux due to this element is given by

$d\varphi =\overrightarrow{B}.\overrightarrow{dA}$

Since magnetic field and area vector are parallel, $\overrightarrow{B}.\overrightarrow{dA}=BdA$

$$\begin{gathered} d\varphi =BdA \\ d\varphi =\frac{{\mu }_{0}Ni}{2\pi r}bdr \end{gathered}$$

To find the flux over the whole cross sectional area, we have to integrate it over r from $r=a$to$r=a+b$ .

$$\begin{gathered} \varphi ={\int }_a^{a+b}\frac{{\mu }_{0}Ni}{2\pi r}bdr \\ \varphi =\frac{{\mu }_{0}Nib}{2\pi r}{\int }_a^{a+b}\frac{dr}{r} \\ \varphi =\frac{{\mu }_{0}Nib}{2\pi r}\mathrm{In}\left(\frac{a+b}{a}\right) \end{gathered}$$

Substituting the given values,

$B=\frac{\left(4\mathrm{\pi }\times {10}^{-7}\mathrm{T}.\mathrm{m}/\mathrm{A}\right)\times 500\times \left(0.800\mathrm{A}\right)\left(0.05\mathrm{m}\right)}{2\times \mathrm{\pi }}\mathrm{In}\left(\frac{15\mathrm{cm}+5\mathrm{cm}}{15\mathrm{cm}}\right)$

On simplifying,

$B=1.15\times {10}^{-6}\mathrm{Wb}$

So the magnetic flux is $1.15\times {10}^{-6}\mathrm{Wb}$