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 :

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

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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