GivenL=1.75mv=50kmh=50000m3600c=13.89mcBr=5×10−5T;SolutionE=BrLv(sinα)=1.215×10−3V(sin90=1)Answer:E(emf)=1.215mVGiven\\L=1.75m\\v=50\frac{km}{h}=\frac{50000m}{3600c}=13.89\frac{m}{c}\\B_r=5\times10^{-5}T;\\Solution\\E=B_rLv(sin\alpha)=1.215\times10^{-3}V(sin90=1)\\Answer:E(emf)=1.215mVGivenL=1.75mv=50hkm=3600c50000m=13.89cmBr=5×10−5T;SolutionE=BrLv(sinα)=1.215×10−3V(sin90=1)Answer:E(emf)=1.215mV
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