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(a) State Faraday's law of electromagnetic induction. (b) Figure 6.60 shows a rectangular conductor PQRS in which the conductor PQ is free to move in a uniform magnetic field B perpendicular to the plane of the paper. The field extends from x = 0 to x =b and is zero for gt b. Assume that only the arm PQ possesses resistance r. When the arm PQ is pulled outward from x =0 to x = 2b and is then moved backward to x = 0 with constant speed v, obtain the expressions for the flux and the induced emf. Sketch the variations of these quantities with distance 0 le x le 2b. |
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Answer» Solution :(a) For Faraday.s laws of electromagnetic INDUCTION, see POINT Number 4 under the heading "Chapter At A Glance" (b) Refer to the figure given with the question LET us first consider the forward motion of the arm PQ from x = 0 to x = 2b. Obviously now the magnetic flux linked with the CIRCUIT SPQR is `phi_(B) = BA`. Hence, (i) `phi_(B) = Blx for 0 le x le b`, and (ii) `phi_(B) = Blb for b le x le 2b.` Consequently the INDUCED emf will `varespilon = - (dphi_(B))/dt.` Hence, (i) `varepsilon = - d/dt(Blx) =- Blv "for" 0 le x le b` and (ii) `varepsilon = - d/dt(Blb) = 0 "for" b le x le 2b` Again for the backward motion of the arm PQ from x = 2b to x = 0, the magnetic flux linked with the circuit SPQR is `phi_(B) =Blb le "for" b le x le2b` and (ii)`phi_(B) = Blx "for" 0 le x leb`.(and flux is gradually decreasing) As a result the induced emf will be (iii) `varepsilon = +Blv "for 0le x le b`. variation of magnetic flux and induced emf during forward and backward and motion of the arm PQ is shown Fig. 6.61. 
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