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Figure shows a conducting rod PQ in contact with metal rails RP and SQ, which are 0.25m apart in a uniform magnetic field of flux density 0.4T acting perpendicular to the plane of the paper. Ends R and S are connected through a 5Omegaresistance. What is the emf when the rod moves to the right with a velocity of 5ms^(-1) ?What is the magnitude and direction of the current through the 5Omegaresistance? If the rod PQ moves to the left with the same speed, what will be the new current and its direction? |
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Answer» Solution :`|e| = Blv = 0.4 xx 0.25 xx 5 = 0.5V` Current , `I = (|e|)/( R) = (0.5V)/(5Omega) = 0.1A` As the rod .PQ. moves to right as SHOWN, the free electrons in it experience a Lorentz force. According to F.L.H., the force is TOWARDS the end .Q. of rod. `therefore `They move from P to Q, hence the end of the rod P becomes deficient of electrons `rArr V_P gt V_Q` APPLYING Fleming.s right hand rule, the current in the rod shall flow from Q to P. (b): If the rod PQ moves to the left with the same speed, then the current of 0.1 A will flow in the rod PQ from P to Q |
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