Question

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

• A. $\frac{B\pi r^2\omega }{2R}$
• B. $\frac{B^2{\pi }^2{r}^4{\omega }^2}{8R}$
• C. $\frac{B\pi r{\omega }^2}{2R}$
• D. $\frac{B\pi r{\omega }^2}{8R}$

Answer 1

The correct option is B $\frac{B^2{\pi }^2{r}^4{\omega }^2}{8R}$

Magnetic flux , $\varphi =\overrightarrow{B}.\overrightarrow{A}=BA\cos \theta =B\left(\frac{\pi r^2}{2}\right)\cos \omega t$

Induced emf , $\epsilon =-\frac{d\varphi }{dt}=B\left(\frac{\pi r^2}{2}\right)\omega \sin \omega t$

Power, $P=\frac{{\epsilon }^2}{R}=\frac{B^2{\pi }^2{r}^4{\omega }^2}{4R}{\sin }^2\omega t$

Mean power, $<P>=\frac{B^2{\pi }^2{r}^4{\omega }^2}{4R}<{\sin }^2\omega t>=\frac{B^2{\pi }^2{r}^4{\omega }^2}{4R}\left(\frac{1}{2}\right)=\frac{B^2{\pi }^2{r}^4{\omega }^2}{8R}$