Week 9 Day 2 Concept Question Solenoid Faraday s Law of Induction Correct answer 3 At time t the gure above right shows a side view of a section of a very long solenoid with radius R carrying current I with magnetic eld pointing up The gure below right shows a top view of the electric eld inside the solenoid at a radius r and the direction of the magnetic eld In the solenoid the current is decreasing in time If C is a stationary closed curve and S is a surface spanning C then The changing magnetic ux through S induces a non electrostatic electric eld whose line integral around C is non zero Lenz s Law Eddy Current Braking Induced current torque or force is always directed to oppose the change that caused it The magnetic induces currents in the metal that dissipate the energy through Joule heating Current is induced counter clockwise out from the center Force is opposing motion creates slowing torque The magnet induces current in the metal that dissipate the energy through Joule heating An in nite straight wire carrying current I is placed to the left of a rectangular loop of wire with width w and length l What is the mutual inductance of the system Correct answer A very long solenoid consisting of N turns has radius R and length d d R Suppose the number of turns is halved keeping all the other parameters xed The self inductance Calculate the self inductance L of a toroid with a square cross section with inner radius a and outer radius b a h and N square windings Current inducted clockwise out from the center Force is opposing motion creates slowing torque EMF proportional to angular frequency Mutual Inductance The current in coil 2 induces magnetic ux F in coil 1 Mutual Inductance M The change in current in coil 2 induces emf in coil 1 Group Problem Mutual Inductance Self Inductance There is self ux LI Faradays law L I t Calculating Self Inductance Assume a current I is owing in your device Calculate the B eld due to that I Calculate the ux due to that B eld Calculate the self inductance divide out I Concept Question Solenoid Group Problem Toroid Inductor Behavior Inductor with constant current does nothing Energy to Change Inductor Start with uncharged inductor Gradually increase current Must do work W LI I Integrate up to nd total work done W LI Energy stored in an Inductor U LI
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