A ring of radius R is rolling on a horizontal plane with constant velocity v. There is a constant and uniform magnetic field, B which is perpendicular to the plane of the ring. Emf across the lowest point A and $$right-most$$ point C as shown in the figure will

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Infinity Learn · Accepted Answer

Point C has two components of velocity: one is along the velocity of ring and other is along the tangent to the ring at that point. That component which is tangent to ring will not create any emf, because it is parallel to the length.Since magnetic field is constant, so circular part of the ring between AC can be replaced by rod AC as shown.So emf across AC $$=$$ BvR

A ring of radius R is rolling on a horizontal plane with constant velocity v. There is a constant and uniform magnetic field, B which is perpendicular to the plane of the ring. Emf across the lowest point A and right-most point C as shown in the figure will

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