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 $ν$ , is

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zero

$\frac{B v π r^{2}}{2}$ and P is at higher potential

$π r B ν$ and R is a higher potential

$2 r B v$ and R is at higher potential

answer is D.

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Detailed Solution

Effective length of the conductor perpendicular to both magnetic fi eld and velocity is 2R.
Induced EMF $= e = BℓV$
Where B = magnetic fi eld perpendicular to the plane going inside.
$ℓ$ = effective length = 2R
v = velocity of the conductor,
So, induced EMF = 2RBv
According to Fleming’s right hand rule, induced current is from P to R.
So, R is higher potential.

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