InterviewSolution
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InterviewSolution
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A square metal wire loop of side 10 cm and resistance 1 ohm is moved with a constant velocity `(v_0)` in a uniform magnetic field of induction `B=2 weber//m^(2)` as shown in the figure. The magnetic field lines are perpendicular to the plane to the loop (directed into the paper). The loop is connected to a network of resistors each of value 3 ohms. The resistances of hte lead wire OS and PQ are negligible. What should be the speed of the loop so as to have a steady current of 1 milliampere in the loop? Given the direction of current in the loop. |
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Answer» Network `AQCS` is balanced Wheatone bridge, no current in branch `AC` `R_(eq)=(6xx6)/(6+6)=3Omega` Resistance of loop `=1Omega` `R_(total)=1+R_(eq)=1+3=4Omega` Induced emf, `e=Bv_(0)l` Induced current `I=(e)/(R_(total))=(Bv_(0)l)/(4)` `10^(-3)=(2v_(0)xx0.1)/(4)` implies `v_(0)=2xx10^(-2)m.//sec=2cm//sec` Flux is decreasing, induced current `I` will be clockwise. |
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