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Question

A conducting straight wire $$PQ$$ of length $$l$$ is fixed along a diameter of a non-conducting ring as shown in the figure. The ring is given a pure rolling motion on a horizontal surface such that its centre of mass has a veleocity $$v$$. There exists a uniform horizontal magnetic field $$B$$ in horizontal direction perpendicular to the plane of ring. The magnitude of induced emf in the wire $$PQ$$ at the position shown in the figure will be :

215145.PNG


A
Bvl
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B
2Bvl
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C
3Bvl/2
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D
zero
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Solution

The correct option is A $$Bvl$$
$$\omega=\cfrac{v}{R}$$
$$=\cfrac{v}{l/2}=\cfrac{2v}{l}$$
$$e=\cfrac{B\omega {l}^{2}}{2}$$
$$=\cfrac { B\left( \cfrac { 2v }{ l }  \right) { l }^{ 2 } }{ 2 } =Bvl$$

Physics

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