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:

  1. (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.