(i) Deduce the expression for the emf induced in the conductor. |
(ii) Find the force required to move the rod in the magnetic field. |
(iii) Mark the direction of induced current in the conductor. |
Answer:
(i) Let the lengths of horizontal arms of circuit be \[{{x}_{1}}\] and \[{{x}_{2}}\] at instants, \[{{t}_{1}}\] and \[{{t}_{2}},\] respectively. \[\therefore \] Area of loop inside the magnetic field, \[{{A}_{1}}=l{{x}_{1}},\] \[{{A}_{2}}=l{{x}_{2}}\] \[\therefore \] \[\Delta \,A={{A}_{2}}-{{A}_{1}}=l({{x}_{2}}-{{x}_{1}})=l\,\Delta \,x\] \[\Delta \,\phi =B\Delta \,A=Bl\Delta x\] \[\therefore \] \[\frac{\Delta \,\phi }{\Delta \,t}=Bl\frac{\Delta \,x}{\Delta \,t}=Blv\] By Faraday's law of induced emf (in magnitude). \[e=\frac{\Delta \,\phi }{\Delta \,t}=vBl\] \[\therefore \] \[e=vBl\] (ii) Current (I) in the loop, Force required must be equal to magnetic force acting on conductor PQ in the opposite directions. \[\therefore \] \[F=IBl\sin 90{}^\circ =(vBl/R)Bl\] \[F=\frac{v{{B}^{2}}{{l}^{2}}}{R}\] (iii) By Fleming's right hand rule, the direction of flow of current is along anti-clockwise direction.
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