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Question
A metallic rod of ‘L’ length is rotated with angular frequency of ‘ω’ with one end hinged at the centre and the other end at the circumference of a circular metallic ring of radius L, about an axis passing through the centre and perpendicular to the plane of the ring. A constant and uniform magnetic field B parallel to the axis is presents everywhere. Deduce the expression for the emf between the centre and the metallic ring.
Solution
Consider the infinitesimally small length dx at a distance x.
So speed of this part is ωx.
Induced small emf = Bωx dx (since emf = vBl)
The emf between the ends of the rotating rod is
`in = int din = int_0^1 Bωxdx = 1/2 Bomegal^2`
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