91Ó°ÊÓ

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

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:

$$\begin{gathered} B=\frac{{\mathrm{μ}}_{0}ni}{2\mathrm{Ï€°ù}} \\ d\mathrm{Ï•}=BdA \end{gathered}$$

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

$B=\frac{{\mathrm{μ}}_{0}Ni}{2\mathrm{π}r}$

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

$d\mathrm{ϕ}=\stackrel{→}{B}.\stackrel{→}{dA}$

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

$$\begin{gathered} d\mathrm{ϕ}=BdA \\ d\mathrm{ϕ}=\frac{{\mathrm{μ}}_{0}Ni}{2\mathrm{π}r}bdr \end{gathered}$$

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

$$\begin{gathered} \mathrm{ϕ}={∫}_a^{a+b}\frac{{\mathrm{μ}}_{0}Ni}{2\mathrm{π}r}bdr \\ \mathrm{ϕ}=\frac{{\mathrm{μ}}_{0}Nib}{2\mathrm{π}r}{∫}_a^{a+b}\frac{dr}{r} \\ \mathrm{ϕ}=\frac{{\mathrm{μ}}_{0}Nib}{2\mathrm{π}r}\mathrm{In}\left(\frac{a+b}{a}\right) \end{gathered}$$

Substituting the given values,

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

On simplifying,

$B=1.15×{10}^{-6}\mathrm{Wb}$

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