"Solution:\\\\ The \\;current \\ is\\ induced\\ in\\ the\\ loop\\;I_i=\\frac{ B}{ t}\\times\\frac{S}{R};\\\\Induction\\ in\\ the\\ solenoid\\;is\\;B=\\mu_0nI_{max},\\\\and\\ then\\ we\\ get\\\\I{max}=\\frac{I_itR}{\\mu_0nS}=\\frac{0.004\\times0.02\\times2}{1.26\\times10^{-6}\\times2500\\times{\\pi}\\times0.5^2}=0.065A;\\\\Answer:0.065A"
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