A conducting bar of length D rotates with angular frequency omega about a pivot

Question

A conducting bar of length D rotates with angular frequency omega about a pivot Pat one end of the bar. The other end of the bar is in slipping contact with a stationary conducting wire in the shape of a circle (only a small part of that circle is shown in the figure). Between point P and the circular wire there is a resistor R. The resistance of the bar and the circular wire are negligibly small. There is a uniform magnetic field B everywhere, it is perpendicular to the plane of the paper. What is the induced current in the loop? Express the answer in terms of D, omega, R, and B.

Explanation / Answer

let us consider a thin section at a distance x of thickness dx.so,

dV=B*v*dx

or dV=B*w*x*dx

integrating with limits from x=0 to x=L,

V=(B*w*L^2)/2

so current = V/R

=(B*w*L^2)/(2*R)

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