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

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

Answer 1

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

Uniform magnetic field of induction $=B$

Radius$=r$

Diameter of the circle$=?$

Frequency$=\omega$

The axis of rotation is perpendicular to the field

Total resistance of the circuit $=R$

Mean power generated per period of rotation$=?$

The flux associated with coil of area A and magnetic induction B is Q

$=BA\cos \theta =\frac{1}{2}B\pi r^2\cos \omega t$

$(\because A=\frac{1}{2}\pi r^2$ for semicircle)

$E_{induced}=-\frac{dQ}{dt}=-\frac{d}{dt}\left(\frac{1}{2}B\pi r^2\cos \omega t\right)=\frac{1}{2}B\pi r^2\omega \sin \omega t$

$\therefore PowerP=\frac{e_{induced}}{R}$

$=\frac{B^2{\pi }^2{r}^4{\omega }^2{\sin }^2\omega t}{4T}$

Hence ${}_{mean}=<P>$

$=\frac{B^2{\pi }^2{r}^4{\omega }^2}{4R}.\frac{1}{2}\because <({\sin }^2\omega t>=\frac{1}{2})=\frac{(B\pi r^2\omega {)}^2}{8R}$