ElectroMagnetic InductionHard
Question
A vertical conducting ring of radius R falls vertically with a speed V in a horizontal uniform magnetic field B which is perpendicular to the plane of the ring :
Options
A.A and B are at same potential
B.C and D are at same potential
C.current flows in clockwise direction
D.current flows in anticlockwise direction
Solution
When the ring falls vertically, there will be an induced emf across A & B (e = Bv (2R)).
Note that there will be a potential difference across any two points on the ring, line joining whom has a projected length in the horizontal plane.
For example, between points ′P′ & ′Q′ there is a projected length ′x′ in the horizontal plane.
∴ P.d. across P & Q is :
V = B v x.
But for points C and D : x = 0.
Therefore; P.d. = 0.
Note that there will be a potential difference across any two points on the ring, line joining whom has a projected length in the horizontal plane.
For example, between points ′P′ & ′Q′ there is a projected length ′x′ in the horizontal plane.
∴ P.d. across P & Q is :
V = B v x.
But for points C and D : x = 0.
Therefore; P.d. = 0.
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