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

Answer 1

Answer: (B)

Magnetic flux $=BAcos\theta =B.\frac{\pi r^2}{2}cos\omega t$
$\therefore {\epsilon }_{ind}=\frac{d\varphi }{dt}=\frac{1}{2}B\pi r^2\omega sin\omega t$
$\therefore P=\frac{{\epsilon }_{ind}^2}{R}\frac{B{\pi }^2{r}^4{\omega }^{2}si{n}^2\omega t}{4R}$
Now, $<si{n}^2\omega t>=½$ (mean value)
$\therefore <P>=\frac{{\left(B\pi r^2\omega \right)}^2}{8R}.$