Electromagnetic Induction - Result Question 4
4. A thin semicircular conducting ring $(PQR)$ of radius ’ $r$ ’ is falling with its plane vertical in a horizontal magnetic field $B$, as shown in figure. The potential difference developed across the ring when its speed is $v$, is :
[2014]
(a) Zero
(b) $Bv \pi r^{2} / 2$ and $P$ is at higher potnetial
(c) $\pi rBv$ and $R$ is at higher potnetial
(d) $2 rBv$ and $R$ is at higher potential
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Answer:
Correct Answer: 4. (d)
Solution:
- (d) Rate of decreasing of area of semicircular
ring $=\frac{dA}{dt}=(2 r) V$
From Faraday’s law of electromagnetic induction
$e=-\frac{d \theta}{dt}=-B \frac{dA}{dt}=-B(2 rV)$
As induced current in ring produces magnetic field in upward direction hence $R$ is at higher potential.
