Question

A thin conducting loop shaped as a semicircle of radius $r$ rotates with angular speed $\omega$ about its straight diameter. The rotation axis is orthogonal to a uniform external magnetic induction $B$. The closed circuit has total resistance $R$. Calculate the average electrical power delivered over one full revolution.

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

Answer 1

Answer: (B)

$\frac{(B\pi r^2 \omega)^2}{8R}$

Answer 2

The magnetic flux through the semicircular loop is given by $\Phi(t) = \vec{B} \cdot \vec{A}(t)$. The area vector of the semicircle is $\vec{A}(t) = A_{semi} (\cos(\omega t)\hat{k} - \sin(\omega t)\hat{j})$, where $A_{semi} = \frac{1}{2}\pi r^2$. The flux is $\Phi(t) = A_{semi} \omega (B_y \cos(\omega t) + B_z \sin(\omega t))$. The induced EMF is $\mathcal{E}(t) = -\frac{d\Phi}{dt} = A_{semi} \omega (B_y \cos(\omega t) + B_z \sin(\omega t))$. The amplitude of the EMF is $\mathcal{E}_0 = A_{semi} \omega B = \frac{1}{2}\pi r^2 \omega B$. The RMS value of the EMF is $\mathcal{E}_{rms} = \frac{\mathcal{E}_0}{\sqrt{2}} = \frac{B \pi r^2 \omega}{2\sqrt{2}}$. The average electrical power delivered is $P_{avg} = \frac{\mathcal{E}_{rms}^2}{R} = \frac{1}{R} \left(\frac{B \pi r^2 \omega}{2\sqrt{2}}\right)^2 = \frac{B^2 \pi^2 r^4 \omega^2}{8R} = \frac{(B\pi r^2 \omega)^2}{8R}$.