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At the instant when the current in an inductor magnitude of the self-induced emf is `0.0640 A//s`, the magnitude fo the self- induced emf is `0.0160 V`. (a) What is the inductance of the inductor? (b) If the inductor is a solenoid with 400 turns, what is the average magnetic flux through each turn when the current is `0.720A`? |
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Answer» Correct Answer - A::B::D `a. L=|e/(di//dt)|=0.016/0.064=0.25H` `b L=(Nphi)/i` `:. =(Li)/N=((0.25)(0.72))/400` `4.5xx10^-4Wb` |
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