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www.GOALias.blogspot.comPhysicsFIGURE 6.11SolutionMethod IAs the rod is rotated, free electrons in the rod move towards the outerend due to Lorentz force and get distributed over the ring. Thus, theresulting separation of charges produces an emf across the ends ofthe rod. At a certain value of emf, there is no more flow of electronsand a steady state is reached. Using Eq. (6.5), the magnitude of theemf generated across a length dr of the rod as it moves at right anglesto the magnetic field is given bydε = Bv dr. Hence,ε = dε=R∫ ∫0Bv drRBωR= ∫ Bωr dr=20Note that we have used v = ω r. This gives2214EXAMPLE 6.61 2ε = × 1.0 × 2 π × 50 × (1 )2= 157 VMethod IITo calculate the emf, we can imagine a closed loop OPQ in whichpoint O and P are connected with a resistor R and OQ is the rotatingrod. The potential difference across the resistor is then equal to theinduced emf and equals B × (rate of change of area of loop). If θ is theangle between the rod and the radius of the circle at P at time t, thearea of the sector OPQ is given by2 θ 1 2π R × = R θ2π2where R is the radius of the circle. Hence, the induced emf is2d ⎡12 ⎤ε = B × R θdt⎢⎣2⎥⎦ = 1 2 dθBωRBR =2 dt2[Note: d θ= ω = 2 π ν ]dtThis expression is identical to the expression obtained by Method Iand we get the same value of ε.