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A metallic rod of length .L. 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 R, 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 present everywhere. Deduce the expression for the emf between the centre and the metallic ring. |
Answer» Solution : The magnitude of the emf, GENERATED across a length dr of the rod, as it moves at right angles to the MAGNETIC field, is given by `d EPSILON = B v dr`. Therefore, `epsilon= ointd epsilon= int_(0)^(R) Bvdr = int_(0)^(R)int B omegardr= (B omegaR^(2))/(2)` |
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