"Solution.\\\\1)\\;the\\; area\\; of\\; a\\; square\\;S=L^2\\\\i=-\\frac{S}{R}\\times\\frac{dB}{dt};\\\\i=-\\frac{L^2}{R}\\times a ;\\\\2)\\;the \\;current\\; is \\;constant, so\\\\P=I^2R\\;orP=a^2R\\\\3)\\;The \\;current\\; flows\\; counter\\;clockwise.\\\\the\\; induction\\; current\\; creates \\\\a\\; field\\; directed \\;against\\; the\\; external\\; field.\\\\4)E_i=-S\\times\\frac{dB}{dt}\\times n;\\\\E_i=-\\frac{\\pi\\times d^2}{4}\\times\\frac{dB}{dt}\\times n;\\\\E_i=-0.0314\\times 10\\times 50=15.7V"
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