Question

In a uniform magnetic field of induction $B$ a wire in the form of a semicircle of radius $r$ rotates about the diameter of the circle with angular frequency $\omega$. The axis of rotation is perpendicular to the field. If the total resistance of the circuit is $R$ the mean power generated per period of rotation is?

Answer 1

Step 1: Given data

Uniform magnetic field of induction$=B$, radius$=r$, angular frequency$=\omega$, resistance$=R$ , the axis of rotation is perpendicular to the field.

Step 2: Solve for the Flux and $E_{induced}$ in the coil

The flux $\left(\varphi \right)$associated with the coil of the area $A$ and magnetic induction $B$ is

$\varphi =BA\cos \left(\theta \right)$

$\Rightarrow \varphi =\frac{1}{2}B{\mathrm{\pi r}}^2\left(\cos \left(\omega t\right)\right)$ (area of semircle=$\frac{1}{2}{\mathrm{\pi r}}^2$, $t=$time taken)

By Faraday's Law induced E.M.F$\left(E_{induced}\right)$

$E_{induced}=-\frac{d\varphi }{dt}$

$$\begin{gathered} E_{induced}=-\frac{d\left({\displaystyle \frac{1}{2}}B{\mathrm{\pi r}}^2\left(\cos \left(\mathrm{\omega t}\right)\right)\right)}{dt} \\ {E}_{induced}=\frac{1}{2}B{\mathrm{\pi r}}^2\sin \left(\mathrm{\omega t}\right) \end{gathered}$$

Step 3: Solve for the Power generated by the coil

Power$=\frac{e_{induced}}{R}$

$$\begin{gathered} =\frac{{\left({\displaystyle \frac{1}{2}}{\mathrm{B\pi r}}^2\sin \left(\mathrm{\omega t}\right)\right)}^2}{R} \\ =\frac{1}{4}\frac{B^2{\pi }^2{r}^4{\sin }^2\left(\mathrm{\omega t}\right)}{R} \\ =\frac{1}{4}\times \frac{1}{2}\times \frac{B^2{\pi }^2{r}^4}{R}\left[min\left({\sin }^2\left(\omega t\right)=\frac{1}{2}\right)\right] \\ =\frac{B^2{\pi }^2{r}^4}{8R} \end{gathered}$$

Hence, the mean power generated per period of rotation is $\frac{B^2{\pi }^2{r}^4}{8R}$.