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State Faraday.s law of electromagneticinduction. Figure shows a reactangular conductor PQRS in which the conductor PQ is free to move in a uniform magnetic field B perendicular to the plane of the paper. The field extends from x=0 to x=b and is zero for x 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 lt x lt 2b. |
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Answer» Solution :The magnitude of the INDUCED EMF in a circuit is equal to the time rate of change of MAGNETIC flux through the circuit. Mathematically, the induced emf is given by ` epsilon = ( - d phi)/(DT)`  First consider the forward motion from `X=0` to `x=2b` The flux `phi_(B)` linked with the section SPQR is `phi_(B) = Blx, 0 le x lt b` `= Blb, b le x lt 2b` The Induced emf is , `epsilon = -(d phi_(B))/(dt)` `= Blv " " 0 le x lt b` `=0 " " b le x lt 2b` 
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