Physics - Magnetism Question with Solution | TestHub

The magnetic field within the turns of toroid is $$\mathrm{B}=\frac{{\mu }_0\mathrm{NI}}{2\pi \mathrm{r}}$$ where, $\mathrm{N}=$ number of turns, $\mathrm{I}=$ current in loops and $\mathrm{r}=$ radius of each turn Here, ${\mathrm{N}}_1=200,{\mathrm{N}}_2=100,{\mathrm{r}}_1=40\mathrm{cm},{\mathrm{r}}_2=20\mathrm{cm}$ and current $\mathrm{I}$ is same, then $$\frac{{\mathrm{B}}_1}{{\mathrm{B}}_2}=\frac{{\mu }_0{\mathrm{N}}_1\mathrm{I}}{2\pi {\mathrm{r}}_1}\times \frac{2\pi {\mathrm{r}}_2}{{\mu }_0{\mathrm{N}}_2\mathrm{I}}$$ Substituting the given values in the above relation, we get $$\begin{array}{rl}\frac{{\mathrm{B}}_1}{{\mathrm{B}}_2}& =\left(\frac{{\mathrm{N}}_1}{{\mathrm{N}}_2}\right)\left(\frac{{\mathrm{r}}_2}{{\mathrm{r}}_1}\right)\\ & =\left(\frac{200}{100}\right)\left(\frac{20}{40}\right)=2\times \frac{1}{2}=1\\ \therefore {\mathrm{B}}_1:{\mathrm{B}}_2& =1:1\end{array}$$