First slide

Magnetic flux and induced emf

Question

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

Moderate

A

zero
B

$B v π R^{2} l 2$ and M is at higher potential
C

$π$ RBv and Q is at higher potential
D

2RBv and Q is at higher potential

Solution

Rate of decrease of area of the semicircular ring $- \frac{d A}{d t} = (2 R) v$
According to Faraday's law of induction induced emf $e = - \frac{d φ}{d t} = - B \frac{d A}{d t} = - B (2 R v)$
The induced current in the ring must generate magnetic field in the upward direction. Thus Q is at higher potential

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