Question

A thin semicircular conducting ring of radius $R$ is falling with its plane vertical in a horizontal magnetic induction $\overrightarrow{B}$. At the position $\text{MNQ}$ the speed of the ring is $v$ then the potential difference developed across the ring is-

• A. Zero
• B. $\frac{Bv\pi R^2}{2}$ and $M$ is at higher potential
• C. $\pi$ $BvR$ and $Q$ is at higher potential
• D. $2BvR$ and $Q$ is at higher potential.

Answer 1

The correct option is D $2BvR$ and $Q$ is at higher potential.


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

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

And $Q$ 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.