Physics - EMI/AC Question with Solution | TestHub

Magnetic flux linked

$=\mathit{\text{BA}}\text{cos}\omega t=\frac{\mathit{\text{B}}\pi {\mathit{\text{r}}}^2\text{cos}\omega t}{2}$

$\therefore$  Induced emf $\mathit{\text{e}}=\frac{-\mathit{\text{d}}\varphi }{\mathit{\text{dt}}}=\frac{-1}{2}\mathit{\text{B}}\pi {\mathit{\text{r}}}^2\omega \text{sin}\omega t$

$\therefore$  Power $=\frac{{\mathit{\text{e}}}^2}{\mathit{\text{R}}}=\frac{{\mathit{\text{B}}}^2{\mathit{\pi }}^2{\mathit{\text{r}}}^4{\omega }^2{\text{sin}}^2\omega t}{4\mathit{\text{R}}}$

                           $=\frac{{\left(\mathit{\text{B}}\pi {\mathit{\text{r}}}^2\omega \right)}^2}{4\mathit{\text{R}}}{\text{sin}}^2\omega t$

$\because$  $<{\text{sin}}^2\omega t>=1/2$

$\therefore$   Mean power generated
                         $=\frac{{\left(\mathit{\text{B}}\pi {\mathit{\text{r}}}^2\omega \right)}^2}{4\mathit{\text{R}}}\times \frac{1}{2}=\frac{{\left(\mathit{\text{B}}\pi {\mathit{\text{r}}}^2\omega \right)}^2}{8\mathit{\text{R}}}$ .