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Physics - EMI/AC Question with Solution | TestHub

PhysicsEMI/ACMotional & Rotational EMFEasy2 minPYQ_2014

PhysicsEasysingle choice

A thin semicircular conducting ring $(P Q R)$ 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:
NEET 2014

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Answer:

Solution:

Here we have to calculate the emf of the conducting ring when it is falling

So, we have to calculate with the following formulae

$e m f = V B l_{e q} = V B (2 R)$

where $R$ is at higher potential and $P$ is at lower potential.

Stream:NEETSubject:PhysicsTopic:EMI/ACSubtopic:Motional & Rotational EMF

⏱ 2mℹ️ Source: PYQ_2014

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