Question

A conducting ring of radius $r$ is rolling without slipping with a constant angular velocity $\omega$ as shown in figure. Maximum EMF induced across any arc subtending angle of ${60}^{\circ }$ at center of ring is

• A. $\frac{B\omega r^2}{2}$
• B. $\frac{B\omega r^2}{3}$
• C. $B\omega r^2$
• D. $\frac{3B\omega r^2}{2}$

Answer 1

The correct option is C $B\omega r^2$

$E_{PA}=\frac{B\omega }{2}{\left(PA\right)}^2=\frac{B\omega }{2}{\left\{2r\sin \right(\frac{\theta }{2}\left)\right\}}^2{E}_{PB}=\frac{B\omega }{2}{\left(PB\right)}^2=\frac{B\omega }{2}{\left\{2r\sin \right(\frac{60+\theta }{2}\left)\right\}}^2{E}_{BA}=E_{PB}-E_{PA}=\frac{B\omega \times 4r^2}{2}\times \left\{{\sin }^2\right(\frac{60+\theta }{2})-{\sin }^2(\frac{\theta }{2}\left)\right\}Formax{E}_{BA},\frac{d{E}_{BA}}{d\theta }=0\Rightarrow \sin \left(30+\frac{\theta }{2}\right)\cos \left(30+\frac{\theta }{2}\right)-\sin \left(\frac{\theta }{2}\right)\cos \left(\frac{\theta }{2}\right)\sin \left(60+\theta \right)=\sin \left(\theta \right)60+\theta =180-\theta 2\theta =120\theta =60at\theta =60°,E_{BA}=\frac{B\omega \times 4r^2}{2}\times \left\{{\sin }^2\right(\frac{60+60}{2})-{\sin }^2(\frac{60}{2}\left)\right\}=\frac{B\omega \times 4r^2}{2}(\frac{3}{4}-\frac{1}{4})=B\omega r^2$