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SC PHYS 202 - Motional Emf

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Phys 202 1nd Edition Lecture 17Outline of Last Lecture I. Ampere’s law and example problemsOutline of Current Lecture II. Motional EmfIII. Example problemIV. SolenoidCurrent Lecture Motional Emf:Emf is the electromotive force and, according to Faridays law of induction, is equal to the change in the magnetic flux over the change in time. The direction of the emf is given by Lenz’s law, which states that the induced current will flow in the opposite direction of the change in flux that produced it.Example problem:A conducting rod is pulled horizontally with constant force F= 4.70 N along a set of rails separated by d= 0.3m. A uniform magnetic field B= 0.600 T is directed into the page. There is no friction between the rod and the rails, and the rod moves with constant velocity v= 5.50 m/s. A) Using Faraday's Law, calculate the induced emf around the loop in the figure that is caused by the changing flux. Assign clockwise to be the positive direction for emf. B)The emf around the loop causes a current to flow. How large is that current? C) From your previous results, what must be the electrical resistance of the loop? (The resistance of the loop is constant.) D)The rate at which the external force does mechanical work must be equal to the rate at which energy is dissipated in the circuit. What is that rate of energy dissipation? x S P These notes represent a detailed interpretation of the professor’s lecture. GradeBuddy is best used as a supplement to your own notes, not as a substitute. x xx xd R QA)ΦB=B(dx) in webersE=-ΔφB/Δt=Bd(Δx/Δt)=Bdv(the negative sign is meant to oppose the increase in magnetic flux) Therefore, current has to be counterclockwise, instead of clockwise. This means that E will be negativeE=.6(.3)(5.5)=.99VB)F=FB=idB4.7=i(.3)(.6)=26.1ACE=V=iR.99V=(26.1)RR=.038OhmsD)Remember that energy dissipation is essentially just powerP=Fv=ΔW/Δt=Δ(Fx)/ΔtP=(4.7)(5.5)=25.85 WattsYou could also use the same equation as we used for power in a circuitPower dissipated=I2R=(E/R)2R=25.8WattsSolenoid:A solenoid is a long, thin loop of wire, wrapped around cylinder in order to produce a uniform magnetic field when an electric current is passed through it. Within the solenoid, the magnitude of the magnetic field is as follows;B=µ0niWhere µ0 is the magnetic constant, I is the current, n is the number of turns per meter(N/L), and I is the current. Outside of the solenoid, the magnitude of the magnetic field is equal to 0. The total flux in the solenoid is equal to the number of turns times B times the area;ΦB=NBa=N[(n/l)µ0i](πr2)E=-ΔφB/Δt=-[(N2/l)µ0πr2](Δi/Δt)E=-L(Δi/Δt)L in this equation is called inductance and the object possessing inductance is an


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SC PHYS 202 - Motional Emf

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