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A horizontal telegraph wire 0.5 km long runing east and west in a part of a circuit whose resistance is 2.5 Omega. The wire falls to g = 10.0 m//s^(2) and B=2xx10^(-5) weber/m^(2) then the current induced in the circuit is |
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Answer» 0.7 AMP Here`df=BxxA=(2xx10^(-5))xx(0.5xx10^(+3)xx5)` dt = time taken by the wire to FALL at GROUND `=(2h//g)^(1//2)=(10//10)^(1//2)=1` sec. `therefore i=(1)/(2.5)[((2xx10^(-5))xx(0.5xx10^(3)xx5))/(1)]=0.02` amp. |
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