Question

A thin semicircular conducting ring$\left(PQR\right)$ of radius $R$ is falling with its plane vertical in a horizontal magnetic induction $\overrightarrow{B}$. The potential difference developed across the ring when its speed is $v$ is

• A. zero
• B. $\frac{Bv\pi r^2}{2}$ and $P$ is at higher potential
• C. $\pi rBv$ and $R$ is at higherpotential
• D. $2rBv$ and $R$ is at higher potential

Answer 1

The correct option is D $2rBv$ and $R$ is at higher potential





As we know, motional e.m.f. , $E=Bvl$

$E=2BvR[\because l=2R]$

And $R$ is at higher potential according to the direction of induced current.

<!--td {border: 1px solid #ccc;}br {mso-data-placement:same-cell;}--> Hence, $\left(D\right)$ is the correct answer.