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The arm PQ of the rectangular conductor is moved from x=0, outwards in the uniform magnetic field which extends from r=0 to x=b and is zero for x > b as shown. Only the arm PQ possesses substantial resistance r. Consider the situation when the arm PQ is pulled outwards from x = 0 to x = 2b, and is then moved back to x = 0 with constant speed v. Obtain expressions for the flux, the induced emf, the force necessary to pull the arm and the power dissipated as Joule heat. Sketch the variation of these quantities with distance. |
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Answer» Solution :Let us first consider the forward motion from X = 0 to x = 2b The flux `phi_B` . linked with the circuit SPQR is `phi_B = Blx"" 0 lt= x lt b = Blb "" b lt= x lt 2b` The induced emf is,`epsi = - (dphi_B)/(dt) = - BLV "" 0 lt= x lt b` When the induced emf is nonzero, the current I is `I = (Blv)/(r )`(in magnitude)  The force required to keep the ARM PQ in constant motion is IlB. Its direction is to the LEFT. ` F = (B^2 l^2 v)/(r ) 0 lt= x lt b` ` F = 0b lt= x lt 2b` The Joule heating loss is `P_f = I^2 r = (B^2 l^2 v^2)/(r ) 0 lt= x lt b` `P_i = 0 b lt= x lt 2b` One obtains similar expressions for the inward motion from x=2b to x=0. One can appreciate the WHOLE process by examining the sketch of various quantities displayed in Fig. |
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