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A metallic rod of length 'I' is rotated with angular frequency of 'omega' 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 its centre and perpendicular to the plane of ring. A constant and uniform magnetic field B parallel to the axis is present everywhere. Deduce the expression for the emf between the centre and the metallic ring. |
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Answer» <P> Solution :Consider a conducting rod PQ of length L, hinged at one end P, and other end Q at the circumference of a circular metallic ring. Let the rod rotates at a uniform angular speed `omega` normal to a uniform magnetic field B.`therefore` Induced emf between the ends of rod i.e., between the centre and the metallic ring `varepsilon=(dphi_(B))/dt =B|(dvecA)/(dt)|` `=Bxx(pil^(2))/T` (where T=time to complete one revolution `=2pi//omega))` 
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