Question

Two thin semicircular ring of radius R/2 are smoothly joined, and is falling with its plane vertical in a horizontal magnetic induction B. At the position MNQ the speed of ring is V. The potential difference developed across the points M and Q is

• A. zero
• B. BV $\pi R^2/2$ and M is at higher potential
• C. $\pi$ RBV and Q is at higher potential
• D. 2RBV and Q is at higher potential

Answer 1

The correct option is A zero

for motion emf between point $MN$

$e=BV(\pi /2+\pi /2)$

$=Bvr$

for motional emf between point $NQ$

$e=Bv(\frac{r}{2}+\frac{r}{2})$

$=Bvr$

Potential difference between two point

$=Bvr-Bvr=0$