A thin semicircular conducting ring (PQR) of radius r is falling with… | NEET 2014 (AIPMT)

Q52NEET 2014 · AIPMTMedium

Motional EMF

A thin semicircular conducting ring (PQR) of radius r is falling with its plane vertical in a horizontal magnetic field B, as shown in the figure.

The potential difference developed across the ring when its speed is $v$ , is

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Answer: (d)

A curved conductor moving with velocity $\vec v$ has the same emf as the straight line joining its ends, taken perpendicular to $\vec v$ .
The ends P and R are a diameter apart: effective length $2r$ , so $\varepsilon = 2rBv$ .
Force on positive charges $q\vec v \times \vec B$ : $\vec v$ downward, $\vec B$ into the page gives a force towards R.
So R is at higher potential. The answer is (d) $2rBv$ , R at higher potential.

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