"Given\\\\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)=1.215mV"
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