Question

A thin semicircular conducting ring of radius R is falling with its plane vertical in a horizontal magnetic field B. At the position MNQ, the speed of the ring is v and the potential difference across the ring is

• A. Zero
• B. $\frac{1}{2}Bv\pi R^2$ 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

Answer: (D)

$2RBv$ and Q is at higher potential

As the ring falls with a velocity v the decrease in area with time is

$\frac{dA}{dt}=-\left(2R\right)v$

$\therefore$ Induced emf, $e=-\frac{d\varphi }{dt}=-\frac{d}{dt}\left(BA\right)=-B\frac{dA}{dt}$

$=2RBv$

From Lenz's law, the induced current in the ring must produce magnetic field in the upward direction. Hence Q is at higher potential.