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 $ω$. 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 $= BA \,cos\,\theta = B. \frac{\pi r^{2}}{2}cos\,\omega t$
$\therefore \varepsilon_{ind} =\frac{d\phi}{dt} = \frac{1}{2} B\pi r^{2} \, \omega \,sin \,\omega t$
$\therefore P = \frac{\varepsilon^{2}_{ind}}{R} \frac{B\pi^{2} r^{4} \, \omega^{2} \,sin^{2} \,\omega t}{4R}$
Now, $< sin^{2}\, ωt > = ½$ (mean value)
$\therefore < P > = \frac{\left(B\pi r^{2}\omega\right)^{2}}{8R}.$