Given:A−cross−sectional area,ρ−resistivity,r−radius of the loop,t−timeSolution:Ei=∣ΔΦΔt∣=∣(0−B)St∣=BSt;I=EiR;R=ρlA;S=πr2;⇒I=BrAtρwithout taking intoaccount the direction of the currentAnswer:I=BrAtρGiven:\\A-cross-sectional\ area,\\\rho-resistivity,\\ r-radius\ of\ the\;loop,\\t-time \\Solution:\\E_i=|\frac{\Delta\Phi}{\Delta t}|=|\frac{(0-B)S}{t}|=\frac{BS}{t};\\I=\frac{E_i}{R}; R=\rho\frac{l}{A}; S=\pi r^2;\Rightarrow I=\frac{B rA}{t \rho}\\without\ taking\ into\\ account\ the\ direction\ of\ the\ current\\Answer:\\ I=\frac{B rA}{t \rho}Given:A−cross−sectional area,ρ−resistivity,r−radius of theloop,t−timeSolution:Ei=∣ΔtΔΦ∣=∣t(0−B)S∣=tBS;I=REi;R=ρAl;S=πr2;⇒I=tρBrAwithout taking intoaccount the direction of the currentAnswer:I=tρBrA
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