An alternating current $\mathrm{I}\left(\mathrm{t}\right)={\mathrm{I}}_0\cos \left(\mathrm{\omega t}\right)$ (amplitude 0.5 A, frequency 60 Hz) flows down a straight wire, which runs along the axis of a toroidal coil with rectangular cross section (inner radius 1 cm, outer radius 2 cm, height 1 cm, 1000 turns) as shown in figure. The coil is connected to a $\text{500\hspace{0.17em}Ω}$ resistor.

In the quasistatic approximation
$\begin{array}{l}\mathrm{B}=\frac{{\mathrm{\mu }}_0}{2\mathrm{\pi s}}\mathrm{\varphi }\\ \text{ So }{\mathrm{\Phi }}_1=\frac{{\mathrm{\mu }}_0\mathrm{I}}{2\mathrm{\pi }}{\int }_{\mathrm{S}}^{\mathrm{b}} -\mathrm{hds}=\frac{{\mathrm{\mu }}_1\mathrm{Ih}}{2\mathrm{\pi }}\ln (\mathrm{b}/\mathrm{a})\end{array}$
This is the flux through one turn; the total flux is N times
$\begin{array}{l}{\mathrm{\Phi }}_1:\mathrm{\Phi }=\frac{{\mathrm{\mu }}_0\mathrm{Nh}}{2\mathrm{\pi }}\ln (\mathrm{b}/\mathrm{a}){\mathrm{I}}_0\cos \left(\mathrm{\omega t}\right).\\ \text{ So }\mathrm{\epsilon }=-\frac{\mathrm{d\Phi }}{\mathrm{dt}}=\frac{{\mathrm{\mu }}_0\mathrm{Nh}}{2\mathrm{\pi }}\ln (\mathrm{b}/\mathrm{a}){\mathrm{I}}_0\mathrm{\omega sin}\left(\mathrm{\omega t}\right)\\ =\frac{\left(4\mathrm{\pi }\times {10}^{-7}\right)\left({10}^3\right)\left({10}^{-2}\right)}{2\mathrm{\pi }}\\ \ln \left(2\right)(0.5)\left(2\mathrm{\pi }60\right)\sin \left(\mathrm{\omega t}\right)\end{array}$
$=2.61\times {10}^{-4}\sin \left(\mathrm{\omega t}\right)$ (in volts, where $\mathrm{\omega }=2\mathrm{\pi }60=\frac{377}{\mathrm{s}}.{\mathrm{I}}_{\mathrm{r}}=\frac{\mathrm{\epsilon }}{\mathrm{R}}=\frac{2\cdot 61\times {10}^{-4}}{500}\sin \left(\mathrm{\omega t}\right)$
$=5.22\times {10}^{-7}\sin \left(\mathrm{\omega t}\right) (\mathrm{amperes}).$
$$\begin{gathered} \begin{array}{l}{\mathrm{\epsilon }}_{\mathrm{b}}=-\mathrm{L}\frac{{\mathrm{dI}}_{\mathrm{r}}}{\mathrm{dt}};\text{ where }\mathrm{L}=\frac{{\mathrm{\mu }}_0{\mathrm{N}}^2\mathrm{h}}{2\mathrm{\pi }}\mathrm{\ell n}(\mathrm{b}/\mathrm{a})\\ =\frac{\left(4\mathrm{\pi }\times {10}^{-7}\right)\left({10}^6\right)\left({10}^{-2}\right)}{2\mathrm{\pi }}\ln \left(2\right)\\ =1.39\times {10}^{-3}\text{ (henries) }\end{array} \\ \mathrm{Therefore} {\mathrm{\epsilon }}_{\mathrm{b}}=-\left(1.39\times {10}^{-3}\right)\left(5.22\times {10}^{-7}\mathrm{\omega }\right) \\ \cos \left(\mathrm{\omega t}\right)=-2.74\times {10}^{-7}\cos \left(\mathrm{\omega t}\right)\left(\text{ volts }\right) \end{gathered}$$